hello good day viewers welcome back once

again today we're going to solve for the

product of X and Y in the following

given Olympia M problem we have here x -

y = 1 and X Cub - y Cub this is equal to

10 in this video I'll be presenting only

one method and this method is awesome

believe me

now for the difference of two cubes

there is a very nice identity that will

be very helpful

here so this identity is X cubed - y

cubed this is equal to x - y

cubed + 3 * x * Y into braet x -

y you know why I'm using this identity

already the value of x - Y is given so

if you plug it in here we're going to

have something really awesome right now

we're going to equate this to be 10

right since we have X Cub - y Cub is =

10 so here we go we have x - y cubed + 3

x Y into bracket x - y this is equal to

10 already we are given that X - 1 is

equal to 1 so we're going to substitute

so here x - y cubed will be the same as

1 cubed

right then + 3 x y MTI X - 1 that is 1

that's x - y I mean then this is equal

to

10 so from here let's simplify this 1

Cub is 1 + 3 x * y * 1 is 3x y right

then this is equal to 10 so let's

subtract one from both sides we get 3x y

this is = 10 - 1 so from here we get 3 X

Y this is = 9 so from here let's divide

both sides by 3 since we're interested

in solving for the value of x * Y and

from here we get the right answer x * y

= 9 3 which is 3 and this is a correct

answer thank you for watching please

kindly subscribe to my YouTube channel

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Olympiad Math | Interesting algebraic problem | #maths #mathematics


hello good viewers welcome back once

value or values of X in the following

given equation we have otk x + x is =

9 so we have here otk x + x this is

equal 9 let's subtract X from both sides

we get root X is equal to 9 - x now

let's go ahead and square both sides

so from here you know the square and the

root sign will cancel out we have here x

is equal to now to expand this let us

recall the following binomial expansion

of a minus B which will give us a 2 + b

2 - 2 * a * B so from here the first

time here square that is 9 Square will

give us 81 then plus the second term

square which is x s

- 2 * the first term and the second term

which is 18 x right now from the right

hand side we have

X2 - 18 x + 81 then from the left we

subtract X right then this is equal to Z

so from here we have x^2 - 18 x - x will

give us - 19 x + 81 and this is equal Z

so from here we're going to solve this

quadratic equation and we're going to

make use of the general quadratic

formula so you know from here this will

be remember a here is the coefficient of

x² which is 1 B is a coefficient of x

which is -19 and C the constant that is

A1 right so here netive B will be

postive 19 plus or minus square root of

B 2 will be -9 s which is same as 19 squ

right because the power is even right so

squ minus 4 * a is 4 * 1 * C which is

the constant 81 then the whole of this

all over 2 * a which is 2 * 1 that is

2 so from here we have X is = 19 plus

orus square root of you know the power

here is even so this will be a positive

result

so that will give us

361 minus 4 * 1 is 4 * 81 that will give

us

324 then all over 2 so this is equal to

19 plus or minus square root of that

will give us 37 right then all over two

so these are the solutions then we have

an exal solution this solution is where

X = 19 +un 37 / 2 so this solution is

extranos you know the reason why I call

it extranos solution because it does not

satisfy the given equation so the only

solution that satisfy the given equation

which we will use therefore X is = 19 -

< TK of 37 all over 2 so this is the

correct solution thank you for watching

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Interesting equation | #maths #mathematics


hello good viewers welcome back once

again today we have this beautiful

equation here and we will be solving for

the values of X we're

given x

+ square < TK of X2 - 9 this is equal to

4 now from here this x here let's move

it to the right so we we are only left

with theare root of x - 9 so here we

have the sare root of x^2 - 9 this will

be equal to 4 - x now from here what do

we do we have root sign here for us to

get rid of this root sign we need to

square both sides of this equation so

here we have the square root of X2 -

9 2 this will be equal to 4 - x^ 2 now

from here obviously the square and the

root sign we cancel out we have

X2 - 9 this will be equal to now from

here let us take note of this

property let us recall that a -

b² this is equal to a

2 + b 2 - 2 * a * B

right okay so from here the first time

here squ will be four Square which is 16

plus the second time here squar which

should be

X2 minus 2 * a b remember 2 * 4 * X

which is 8 x right so we can see the way

we easily expand this right so here we

have

X2 now we have - 9 here then this is

equal to let's write this as

x² - 8 x then + 16 right okay now we can

see from both side x squ is both

positive right so which actually mean

they will cancel out so but this is how

we do this let's move this whole term to

the left so here we have

X2 - 9 here this is positive so it will

become negative so we have - x^ 2 this

one is negative so it will become

positive + 8 x this one right over here

is positive so it will become negative

minus 16 and this will be equal to zero

so from here we can see x^2 - X2 is 0 we

have here 8

x now -9 - 16 we give

us 25 right yes of course so that is 25

then this is equal to 0 now this 25 is

negative if it's crossing over it will

become positive so we have 8 X is = to 0

+ 25 will give us 25 since we're solving

for x let's divide both side by the

coefficient of x coefficient of x is

this number right over here attached to

X so here we have 25 / by 8 this cancel

this and here we have our solution that

x isal 25 / 8 thank you for watching

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Interesting equation | Can you solve ?


hello welcome back here we have another

interesting Olympiad match so but this

is Jun

Olympiad here we given the following

equation x + 1 x is = 3 so we are asked

to evaluate X2 + 1 /

X2 obviously solving for the value of

values of X in this equation before

evaluating this seems not to be the best

way to solve this problem this problem

must have an easier way to deal with

it okay so now let us see how we're

going to find the value of X2 + 1 / X2

without first figuring out the value of

values of X in this equation right okay

now let us recall this algebraic

identity we have that a s + b 2 is equal

to a + b 2 - to a right so this identity

is going to help us so much we're going

to use this identity to find the value

of X2 + 1 / X2 which that first solving

for the value of values of X in this

equation so now we can see here we have

a 2 + b s and we have similar expression

here which have power two and here power

two you know 1 / x² is the same as 1 /x

then everything squared right okay now

let us reite this expression so we're

going to have here

x² +

1 /

X2 to be equal to you know here we have

a so here we're going to have X here

then plus you know the whole of this

thing is squared just like I showed here

so this is going to be here 1

/x then squ minus here we have 2 * a so

here a is X here right so we're to have

X then times here we have 1 X you know a

here a so we're going to have 1 x here

right so we can say that this cancel out

right so this is going to be X2 + 1 / X2

to be equal to x + 1 /x 2 then - 2 right

so this is going to be equal to now what

is the value for x + 1 / X so we have it

right over here so this is going to be 3

S then - 2 you know 3 S is 9 then we

have - 2 and this is equal to 7 so

therefore the value for X2 + 1 / X2 is 7

thank you for watching please do well to

follow this page subscribe to my YouTube

channel and share this video

Interesting Junior Olympiad Math


hello welcome back once again please

subscribe to my

Channel please like comment and share

here we're giving the

following equation right over here 8^ of

n + 8^ of n + 8^ of n all 4^ n + 4^ n

this is equal to 48 and right here we

asked to find the value of n sorry for

the X here

right okay so here we have a ^ n + a^ n

+ a^ N All 4 the power of n + 4 the ^ of

N and this is equal to

48 okay mind you here we have x 8 to^ n

+ 8^ n + 8 ^ of n let us recall that

when we have x + x you know this is

equal to 2x we have x + x

+ x this is equal to 3x that is 3 * X

right so right over here we have a^ of n

+ a^ of n + a^ n so this going to be 3 *

^ of n right so 3 * 8^ n and then here

we have 2 * 4^ n right then this is

equal to

48 okay now we're going to express8

an index Power of Two And also four we

all know 8 is = 2 cubed and 4 is = to S

right so here we're going to have three

times right over here we have 2 cubed of

n then all over 2 * here 2^ 2 of n this

is equal to

48 so according to the law of indices

we're going to multiply these Powers

together so right over here we're going

to have three * 2^ 3 n all over 2 * 2^ 2

N we multiply this Powers together and

this is equal to

48 okay so from

here let us multiply both sides by two

so here we multiply by two and here we

multiply by two so these two cancel out

this two and from here we have equal

bases so we're going to subtract the so

here we have 3 * 2^ 3 N - 2 N and this

is equal to 48 * 2 so let us go ahead

and divide both sides by three so here

by three and here by three right okay so

here three cancel three and right over

here we have 2 of 3 N - 2 N that is n so

this is equal to you know here three

here 1 and three here is 16

right so right over here we have 16 * 2

so that is going to give us 32 and this

32 is the same as 2 to^ of 5 so from

here we have 2 ^ of n is = 2 the Power

of 5 so we can see we have equal bases

so we can easily equate the powers and

therefore we find the value of n to be

equal to five thank you for watching

please kindly subscribe to my Channel

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Interesting exponential equation.


good day viewers welcome back to my

Channel today we have the following

limit problem limit as X is approaching

one the root of x - 1 x -1 was touch wow

I can say that this limit is extremely

beautiful because of the way the limit L

wow now I'm going to start by making a

substitution so let squ otk of x = k

Square both side we obtain x = k two if

our X is approaching one k is also

approaching one so our limit becomes

limit as K is approaching one here we

have Kus one all over you know here k 2

- 1 okay now we using the difference of

two squares formula that a 2 - b² is

equal to the product of a + b and a

minus B okay so our limit

becomes limit as K is approaching 1 up

here we have kus1 all over down here K

-1 and k + 1 so K -1 gets R kus1 so we

have what limit K is approaching one of

1 k + 1 and is equal to 1 1 + 1 and is

equal to 1 over 2 which is our desired

answer thank you for viewing

please subscribe to my YouTube channel

for more interesting videos

A very nice and interesting limit. Avoid L'hospital's rule !!!


hello welcome back once again today we

have this olymp mathematics problem

we're given that sare root of A+ square

root of B is = 2 a * B is = 4 so here we

be evaluating a + b right so from

equation one we have root of

A+ square root of B which is equal to 2

now let's go ahead and square both sides

of this equation okay so from the left

hand side let us make use of this

property we know a + b 2 is = to a 2 + b

2 + 2 * a * B so from here we're going

to have the first term in the bracket

squar which is root a 2 plus the second

term which is root B

2+ 2 MTI the first term of a * the

second term root of B and this will be

equal to 2 s from the right hand side

which is

4 from here we know Square will cancel

square root square will cancel square

root so we left with a plus b then plus

we have here 2 now root a * root B will

be same as square root of a * B under

one root sign then this is equal to 4

now let us recall back we know a * B is

= 4 so therefore we're going to

substitute the value of a * B here so we

get a +

B+ 2 square < TK of 4 right because a *

B is 4 this will be equal to 4 so here

we have a + b + 2 * > 4 is 2 then this

is equal to 4 from here we get a + b + 2

* 2 is 4 which isal to 4 now subtract

four from both sides of this equation so

we have a + b + 4 - 4 this will be = to

4 - 4 so from here this is zero right so

we get our right answer to the problem A

+ B = 4 - 4 which is 0 and this is the

right answer than thank you for watching

please kindly subscribe to my channel

also like comment and share bye-bye

Olympiad Math | Can you solve ?


hello welcome back once again today

we're going to solve this nice equation

so from the left hand side we have this

square root of we know X Cub + x Cub is

2 X Cub right and from the right hand

side this is equal to x^ 2 now we're

going to square both sides to get rid of

this root sign so let's Square both

sides of the

equation so we know the square we cancel

the square root

so leaving us with 2 x Cub so this is

equal to you know 2 * 2 is 4 so that

would be x to^ 4 right okay now let us

make the right hand side the left hand

side and the left hand side the right

hand side so we now have x^ 4 is = 2 X

Cub right so here subtract 2 x Cub from

both side so this will be x^ 4 - 2 X Cub

this is equal Z

from the left hand side x Cub is common

so let's Factor it out so we're going to

have x^ 4 x Cub will give us x - 2x Cub

/ X Cub will give us 2 and this is equal

to Z so we have these two products which

is equal to Z so we have either X cubed

is equal to 0 or x - 2 is = 0 so from

here take the Cub root of both side we

get get x = 0 and this is one of our

solution and from here add two to both

side we get x = 2 so these are the only

two possible real solutions that satisfy

the given equation thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Canada | Nice Olympiad Math problem | can you solve ?


hello good viewers welcome back once

again please kindly subscribe to my

channel also like comment and share

today we have this nice exponential

equation problem we're going to solve

for x we are given that 4 to the^ of x +

4 ^ of X is = to 8 here we're going to

solve for the value of x right okay

let's get started we have 4 ^ x + 4 ^ x

this is equal to 8 so from the left hand

side we're going to factor out 4^ of X

and here we have 4^ x / 4^ X we have 1

plus also here we have 1 and this is

equal to 8 right okay so from here 1 + 1

is 2 so here we have 2 * 4 ^ of X this

is equal to 8 so from here let's go

ahead and divide both sides by two so

divide this so so here we get 4 to the^

X is equal to you know 8 / 2 that is 4

right so here you know this four is the

same as 4 to the^ of 1 and from here we

have equal bases can actually equate the

powers and this gives us X isal to 1 and

that is the right answer thank you for

watching please do wa to subscribe to my

channel also like comment and share

bye-bye

A very simple exponential equation. For beginners, here is step by step on how to solve this problem


hello good viewers welcome back once

again here we have another interesting

algebraic problem here we're given that

A + B is = 10 and right over here we ask

to evaluate a - 7 ^ 7 + b - 3^ of 7

right okay now let's get started from

this equation right over here we have

that A + B is equal to 10 so from here

we can solve for a in terms of B that is

making a the subject of formula here so

we're going to have a is equal to 10 - B

right okay in the given expression right

over here we are going to substitute the

value for a in terms of B so here we

have a -

7^ 7 + b - 3^ of 7 so this is equal to

so here we have a that is 10 - b - 7 ra

power of 7 then plus b - 3 power of

7 right okay so from here we own that

here we

have we have here 10 - 7 right so to

group it then minus B 7 + b - 3 7 so

this is equal to here here we have 7 10

- 7 which is 3 right so here we have 3 -

B 7 + b - 3^

7 okay so from here this is equal to you

know here we can Factor

out1 right so let us factor out1 from

here so here we're going to open a

bigger bracket so factor out -1 then

Open Bracket you know this is going to

affect what we have inside this bracket

so it will become b - 3 right then the

whole of this of 7 then plus b - 3 7 so

this is equal to here we're going to

make use of this law so when we have a *

m of n this is same as a of n * m of n

right so from here this going to be -1

of 7 then time b - 3 7 right then plus

here b - 3^ of 7 so from here this is

equal to you know 1^ of an odd power

will give us -1 so we have Open Bracket

b - 3 7 + b - 3^ 7 and from here you

know netive B - 3 7 + b - 3 7 will give

us zero and here comes the right answer

to the question thank you for watching

please do wait to subscribe to my

YouTube Channel please like comment and

share

Olympiad Math problem. How to evaluate?


hello welcome back once again today we

have this Olympiad maths we have a 2us

199 a =

199 so here we're going to evaluate this

expression here a ^ 4 + a 2 + 1 / a cub

1 first of all let us write this

expression in the simplified

form the quatic expression here can be

factorized so I think I will discuss

that in my next video when you factorize

this we're going to

have a 2 + a + 1 mtip by a² - A +

1 right then divided by so the

denominator here we're going to make use

of the sum of two cubes you know we have

a cub + 1 which you can write as 1 cubed

so this will be equal to

recall a cub + B Cub = A + B * a 2 - A B

+ b 2 so make use of this of this sum of

two cubes formula so we get a + 1 * a 2

- A + 1 s which is 1 right so here we

get a + 1 1 * a 2 - A + 1 so of which

this we cancel this and here we have a 2

+ a + 1 / a + 1 now let's split this

denominator as follow here a

2 / a + 1 then + a + 1 / a +

1 so this will cancel is living one

right so here we

get

a² / a + 1 + 1 right so this is the

simplified form of this expression right

over

here okay now from the original given

equation remember we have

a² - 199 a

this equals

199 so now this term here which is

negative let's move it to the right hand

side and that will become positive so we

get a² =

199 + 199 a right so here

remember is common right so we're going

to factor it out so we get a² = 199 into

bracket here will be 1 and here will be

a which is same as a + 1 right so here

the value for a s we're going to

substitute it here so this will be equal

to here instead of a s we use that so

that is 199 into bracket a + 1 then

remember we're dividing by a + 1 then +

1 so this equals here this cancel this

and here we get 199 + 1 which is equal

to 200 which is the right answer thank

you for watching please can subscribe to

my channel also like comment and share

bye-bye

Germany | Nice Olympiad Math | Find out the trick in this video


hello good viewers welcome back once

again let's find a solution to this

exponential equation we have the X root

ofx this is equal to

27 before we proceed it will be very

necessary to make a substitution

here we can

let x = - y now when you multiply both

side by -1 we also getx is = 1 y now the

left hand side of this equation we can

rewrite it make use of this identity the

mot of n is same as n^ 1 / n so from the

left hand side we

havex to the^ of 1 /x and this is equal

to

27 now making use of the

substitution X is

y then ra power of 1 over X which is

what - Y and this is equal to 27 from

here we get X sorry I mean y of -1 / y

this is equal to

27 now the power of Y here we can write

it as -1 * 1 / Y and this is equal to 27

now from here we're going to make use of

this property a a power of B * C this is

same as a power of B in bracket then

power of C so from here this becomes y^

of1 of Y 1 / y I mean this is equal to

27 and you know this is the reciprocal

of Y which is the same as 1 / y^ of 1 /

Y and from the right hand side 27 is 3

of 3 now from here this implies that 1 /

Y is = to 3 this is over one when we

cross

multiply we get 3 Y is = 1 and from here

we get Y is = 1 /

3 okay now let us

recall let us recall that X is = y so

from here we come up with the solution

that X isal

to1 /3 and this is the right answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Indians | Nice exponential equation for Olympiad | #maths #mathematics


hello welcome back once again today

we're going to solve this radical

equation here we have the square root of

Cub root of x this is equal to the Cub

root of x here we're going to solve for

real possible values of X right now

let's get started let us take note of

this generally the end

root of a is equal to a

of 1 / n right so therefore the square

root of let's say k will be same as K to

the^ of 1 / 2 right so from the left

hand side we have the square root of

cube root of x so we're going to write

it

as the cube root of

x then R of 1 / 2 then this will be

equal to from the right hand side make

use of these algebraic properties so we

have here x ra power of 1 / 3 right okay

now do not forget here in this base here

we still have the CU root of x so we can

which we can actually re write again as

x^ of 1 / 3 then in bracket ra the^ of 1

/ 2 right then this will be equal to X

to the^ of 1 / 3 now take note of this

again when we have a to the power of M

then in bracket to the power of n this

is same as a m * n right so from here we

have here x 1 3 * 1 / 2 this is 1/ 3 * 1

/ 2 so this will give us 1 / 6 so right

over here we have 1 / 6 then this is

equal to X to the^ of 1 / 3 right okay

now from here let us check the LCM of 6

and three we all know the LCM of 6 and

three or 3 and 6 will be equal to 6

right so here we're going to raise both

sides to the power of six so here we go

we have here x^ 1 / 6 to^ 6 this will be

equal to x^ of 1 / 3 to the^ of 6 so

from here make use of this property here

this particular one here so this is

going to be 1/ 6 * 6 is 1 right so we

are left with X here then from the right

6 * 1 / 3 will give us 2 right so from

the right hand side we have x²

then this is equal to from the left hand

side we have only X okay so from here

subtract X from both sides we get x^2 -

x this is equal to Z so from here x is

common so we Factor it out and we have

here x -1 this is equal to Z right okay

so from here when we have these two

product here and they are equal to zero

so we have either this x here is equal 0

right and this is a solution or x -1 is

0 of which this implies that X is equal

to 1 so here we have the only two R

solutions that satisfy the given radical

equation thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting radical equation| Can you solve ?


hello welcome back once again today we

have beautiful systems of equations we

have x * y * Z 2 x + y + z = 6 x y * y z

+ x z = 11 you going to evaluate x / y z

+ y / x z + z / XY right okay from

equation 2 we have x + y + z = 6 let's

go ahead and square both

sides so from the left here this is

exactly equals X2 + Y 2 + Z 2 then plus

the other terms we have 2 * 2 * 2 * so

we can just factor out 2 and we get x y

+ y z + x Z then this equals 36

okay so from here let's solve for X2 + y

+ Z 2 so we have here X2 + Y 2 + Z 2

here plus here x * y * x * y + y * Z + x

* Z = 11 right so this will be 2 * 11

which is 22 and this will be

36 so from here

x2+ y 2 + z^ 2 this = 36 -

22 36 - 22 that will give us 14 right

okay now take a look at the expression

we asked to

evaluate now have you thought of

something like this

X2 + Y 2 + Z 2 when you divide this by x

* y * Z splitting the denominator we get

x2 / x * y * Z + y^ 2 / x * y * Z + Z 2

/ x * y * Z which is equal to here x

X2 will be reduced by X here so we get X

all over y z right so we get here x over

y z and plus from the other side that

will be y over that is x z plus the

other side will be Z all over X Y wow

and we can see this is exactly the

extion we ask to evaluate but the

question here is what exactly yelds this

here it is so this expression right of

over here equals this so our answer will

be the value of this ID by the value of

this and the value of x + y + Z2 = 14 ID

by the value of x y z given in the first

equation equals 2 so here we found the

solution which is s and here is the

right answer thank you for watching

please kindly subscribe to my channel

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Interesting Olympiad Math


hello welcome back once again today we

have this nice Olympiad mathematics

problem we have x + 1 /x is = tok3 so

here we be evaluating x -1

/x now let's get into the video from the

given equation we have x + 1

/x which is equal to < tk3 now let's go

ahead and square both sides of this

equation

now from the left hand side here let us

make use of this property a + b s this

is equal to ter of aus B so that will be

a - b 2 + 4 * a * B so make use of this

property at the left hand side so that

will become x -1

/x then 2 + 4 *

x * 1

/x then this will be equal to here

Square we cancel root sign so we have 3

okay so from here we have X - 1

/x 2 then plus here x cancel X here so

we have four then this is equal to 3

subtract four from both sides we get X -

1 / x^ 2 this will be equal to 3 - 4

which is -1 take root of both sides and

from here we have our Solution that's x

-1 / X is equal to plus or minus theun

of1 and this is plus or minus the

imaginary unit I right so our solution

for x -1 /x is = to Plus or minus Iota

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting Olympiad Math | Can you solve ?


good day viewers welcome back to my

Channel today we have the following

interesting limit problem limit as X is

approaching 0 1us the cine of X all 2x^

2 = that so we're having a restation

here so we are not going to make use of

the L hital rule in this case so this is

not allowed okay so let's how we're

going to solve this problem okay so I'm

going to use the idea of the power

series expansion of the cosine function

okay so I'm going to do it as

follows COS of X is = to 1 -

x 2 / 2 factorial + x 4/ 4 factori minus

and so on okay so so from here we can

see that 1 - the cosine of x is the same

as 1 - 1 - x 2 / 2 factorial + x 4 / 4

factorial minus and so on so by so doing

we get this equals so you know opening

this bracket with this negative sign so

1 - 1 is 0 then we have x^ 2 / 2

factorial - x 4 for factoral plus and so

on okay so and this is equal to 1us the

cosine of x so dividing both side by 2

X2 okay here over 2 x² so we get 1 - the

cosine of x all/ 2 x 2 is = to you know

here x² cancel x s here so we get 1 / 2

factorial /

2 minus you know x² reduces this to x 2

/ 4 factoral over 2 plus and so on okay

so from here we can see that the

limit as X is approaching Z of 1us the

cosine of x all 2 x^2 is the same as

limit as X is approaching Z of 1 / 2

factorial 2 - x^2 / 4 factorial / 2 plus

and so on okay so from here we can see

that as X is approaching zero the whole

of the terms here that goes on to

Infinity will be approaching zero so we

are left with 1 all 2 Factor all over 2

and this equal to 1 / 2 / 2 so this is

equal to 1 4 so our limit equals one all

over four thank you for viewing and

please make sure you subscribe to my

YouTube channel for more interesting

videos

Limit. Can you do this without L'hospital's rule? Check out the video.


hello welcome back once again today

we'll be

evaluating the four power of < TK of 45

+ 5un 5 all

/ < TK 20 okay let's get started first

of all we need to take note of this we

know Square < TK of

45 is equal

to square < TK of 45 is the same as

9 * 5 right so this will be equal

to square < TK of 9 then multip

by square < TK of 5 right so this is

equal to root of 9 is 3 so we have 3 <

TK 5 right okay and again we have square

root of 20 right over here

so we know sare < TK

20 is equal

to square < TK of

4 *

5 and this is equal

to < TK

* < TK

5 we know otk of 4 is 2 so this becomes

2un 5

okay so this whole expression here will

become so this whole expression

here will

become so that will become in

bracket 3 < TK

plus

5 < TK 5

ID

by 2 < TK 5 then all rais the power of

4 okay so this will be equal to into

braet here 3un 5 + 5un 5 will give us 8

< TK 5 right then divid by 2un

5 ID by 2 < TK 5 all to the^ of 4 so

this will be equal to we know < TK 5 we

Cancun 5 from here and 8 / 2 is 4 so

this will be 4 ra^ of 4 right so it is

easy for one to know that 4^ 4 means 4 *

4 *

4 * 4 right which is equal to 16 * 16

because from here this is 16 and here

this is 16 okay and 16 * 16 is easily

gotten as

256 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Interesting problem | No calculators !!!


hello welcome back once again to my

YouTube Channel please kindly support

this channel by

subscribing please subscribe to my

YouTube channel and share my videos

please subscribe here we are asked to

evaluate this expression right over here

586 s - 16 all divide by

590 right so let's get started

here we have

SAR -

16 all over

590 so this is equal to

586 squ minus you know 16 is the same as

squar all/ by

590 okay before we proceed let us recall

this difference of two squares formula

a

2 - b 2 is = to a + b multi a minus B

right so we're going to make use of this

property here so this is going to be

equal to in the first bracket here we

have a s and here we have a so here we

have 586 squ so the first bracket we

have

586 then plus b that is the base here

then the base here is four so we have +

4 MTI by

586 -

4 all divide by

590 so this is equal to you know 586 + 4

will give us

590 *

586 - 4 will give us

582 right all divide by

590 so we can see that this 590 we

cancel out so we are left with

582 and this is the right answer please

and please subscribe to this YouTube

channel and share our videos remember to

subscribe thank you

The difference of two squares is a way out.


hello welcome back once again today we

have this interesting equation involving

cube root and fourth root so here will

be soling for the real values of X right

now let's get started let us take note

of this property we know the end root of

a that is generally this is equal to a

to the^ 1 divid n so therefore the cube

root of x will be

let's put it down first we have cube

root of x which is equal to the fourth

root of x so with this identity here

cube root of x will become x^ 1 / 3 then

this will be equal to 4 root of x will

be x^ of 1/ 4 right okay now we're going

to raise both sides of this equation to

a specific power and how do we know the

power to raise them two right we need to

get rid of this cube root and default

root so we can get a standard equation

to solve okay so what are we going to do

let us look for the LCM of 3 and four we

know the

LCM of 3 and 4 is equal to 12 right so

we're going to raise both sides of this

equation to the power of 12 so here we

have X to the^ 1 / 3 to the^ of 12 this

will be equal to x^ 1 4^ 12 we know this

property a^ b the^ c is equal to a^ B *

C so with this property we have here x^

1 3 * 12 is as 12 / 3 then this will be

equal to x^ 12 * 1 4 is 12 / 4 so we can

see this can divide nicely right so we

have here x^ 12 3 is 4 then this is

equal to x^ 12/ 4 is three so we have

this standard standard quartic equation

to solve so let's move this term to the

left hand side so we have x to^ 4 - x

Cub this is equal to0 so many may think

that the best way to solve this equation

from here is to divide both side by x^ 3

my dear if you do that you're going to

lose a solution of x is equal to Z which

is very true right so the best way to

solve this kind of equation is to move

the term from the right to the left and

the sign will change right and from here

we can see x^ 3 is common so let's

Factor it out we have x^ 3 into bracket

x^ 4 / X Cub will give us just

xus x^ 3 / x^ 3 is one and therefore we

have this equal

Z okay so from here we have these two

products which is equal to 0 so we have

either x^ 3 is = 0 or we have X - 1 is =

0 so take root of both side we get X

isal to 0 so this is a solution from

here add one to both sides we get the

final solution X isal to 1 and here we

have the two real solutions that satisfy

the given interesing

equation thank you for watching please

kindly subscribe to my channel so like

comment and share bye-bye

Interesting radical equation | Can you solve ?


hello good day viewers welcome back once

again please kindly support this by

subscribing also like comment and share

today we're going to find the derivative

of x to the^ of X right with respect to

X here we have y is = X to the power of

X

right so here we're going to take the

natural log of both sides so here we

have the natural log of Y this is equal

to the natural log of x to the power of

X now with one of the properties of log

gring that is the natural log of a to

the^ B is equal to B MTI the natural log

of a so here we're going to have the

natural log of Y is equal to x * the

natural log of x right so from here

we're going to differentiate now by the

right hand side we're going to make use

of the product rule right so when we

have the derivative with respect to

X of let's say we have

ax time BX right so this is equal

to a x Prime means derivative of ax *

BX then plus BX Prime is the derivative

of BX * a x

right okay

you know here ax and BX are different

functions of X

right okay so now here we're going to

have the derivative of natural log of Y

this is going to be 1 over y so did not

forget here we're going to have then d y

DX right so this is equal to here we're

going to have the derivative of x here

according to this up to here the

derivative of x here is 1 then times

natural log of x right then here

plus the derivative of the natural log

of x is 1

X then times x right making use of this

property

right okay from here we're going to

simplify here we have 1 / y time the Y

the

X this is equal to here we have the

natural log of x plus here this this so

we have here one right so from here

we're going to multiply both side by Y

and here we get d y DX is going to be Y

into bracket the natural log of x + 1 so

do not forget that Y is exactly equal to

x^ of X so here we get x to the of X

into bracket the natural log of x + 1

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Differentiation. The derivative of  x^x. #education #maths


hello good viewers welcome back once

again let's find a solution to this

Olympiad mat problem we're given that X2

+ 1 x² is = to root of 13 right over

here we're going to evaluate this

expression x² - 1

/x2 now let's get started before we

proceed there is a very nice identity

we're going to make use of remember that

A+ b 2 this is same as a - b 2 + 4 a

b okay now from equation one we have X2

+ 1 /

X2 this is equal to root of 13 now let's

go ahead and square both sides of this

equation so that we can make use of this

identity now from the left side if you

make use of this identity we have here

X2 - 1 / X2 then

squ + 4 * a here will be

X2 * 1 / x² right then this is equal to

you know here the square and the root

sign we cancel out and we're left with

just

13 now from here don't forget that x²

and 1 / X2 we cancel out and here we

have X2 - 1 / X2 2 then + 4 this is

equal to 13 now let's go ahead and

subtract four from both side we get from

here x^2 - 1 / x^2 2 this is = to 13 - 4

which is 9 now from here let's take root

of both sides and here we get x2 -

1 / X2 this is equal to plus orus otk 9

and from here this is exactly equal to

plus or minus 3 and this is the right

answer thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Ukrainians | Nice Olympiad Math | Here is the trick | #maths #mathematics


hello welcome back once again today we

have this interesting exponential

equation and here we're going to be

solving for the values of X right so

let's get started so from the left hand

side we have

9 of < TK x +

9 < TK X and this isal to

54 so from the left hand side we can see

that 9^ ofun X is common so let's Factor

it out so here we have 9 ofun x into

bracket so 9 ^un x / 9^ root X is

1 plus 9^ root x / 9^ root X is 1 then

this is equal to

54 so in our next step we can see that 1

+ 1 here is 2 so here we have 9 to the

of otk x * 2 then this is equal to

54 so from here let's divide both side

by two so divide this by two and divide

here by two so two cancel two here then

here we have 9^ ofun X is equal to 54 /

2 that will give us

27 right so now we can express 9 in

index power of 3 as 3^ 2 then we have

here to the^ root X this is equal to 27

is same as 3 to the^ 3 now let us apply

this rule we know that A to the^ B to

the power C isal to a^ B * C so from

here this will be 3^ 2un X this is equal

to 3 to the^ 3 so from here we can see

that we have equal base from both sides

so we're going to equate the powers so

this implies from here that 2un x isal 3

now divide both by

two this two will cancel this two so

here we get root X is equal to 3 / 2 now

for us to get rid of this root sign

let's Square both sides of this equation

so by squaring both sides Square will

cancel square roots and here we have our

solution X is equal to 3 S here is 9 / 2

2 is 4 and this is the correct answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation | can you solve ?


hello good day viewers welcome back once

again today we have the following

interesting quartic equation x ^ 4 + 2 x

Cub + 4 X2 + 8 x + 16 this is = z right

over here we're going to evaluate x to

the power of

10 now there is a very interesting

algebraic Identity or let me call it

formula we're going to make use of

here let us

recall that a to the^ of

5- B to the^ of 5 can be factorized as a

- b then multi by a to the^ 4

plus a cubed MTI B plus we have here a 2

* b 2

then

plus a * B Cub + B to the power of

okay

now let a equals

X

and let b equal

2 now we derive the following identity

in terms of X that would be a ^ 5 will

be x ^ of 5 minus b ^ 5 will be 2 ^ of 5

then this is equal to a - b will be x -

2 right then multiply by a to ^ 4 will

be x ^ of 4 + a cub B will be 2 * X Cub

right then plus a 2 b 2 will be 2 2 * x

2 which is 4 x²

[Music]

plus a * B Cub would be x * 2 Cub which

is 8 x then plus b^ 4 is 2^ 4 and that

is

16 okay now carefully looking at this

equation from

here we will realize that at the right

hand side this quatic expression here is

exactly equal to zero just like it is

given right over here here so we're

going to substitute so we get x ^ 5 - 2^

5 this is equal to x - 2 then multi by

0 now from here we get x ^ 5 - 2 ^ 5

this is equal to this * 0 will give us 0

from here we move 2 the^ 5 to the right

so we get x the^ 5 this is equal to 2^

of

5 now let us recall that we are asked to

evaluate x^ of

10 now let us raise both side to the

power of

two as we know this property a^ of B in

braet C this is equal to a of B * C so

from here we have x ^ 5 * 2 which is 10

and this is what we ask to find this is

equal to 2 ^ 5 * 2 which is 10 and

simplifying this this is exactly equal

to 10 to 4 and this is the right answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Japanese | Nice Olympiad Math | Interesting problem | #maths #mathematics


hello welcome back once again today we

have this interesting

equation involving powers of logorithm

and here we'll be solving for the value

of x now anytime we see a logarithmic

function without a specific base then we

need to know that the base must be

10 right now let's start with the

following substitution we're going to

let log X

M now remember this base here is 10

right so we know the rule 10^ of M will

be equal to X right so we have X is =

10^ of M now from here let's go back to

original equation and make the

substitutions so here x as the base here

will be 10 the^

n then we need to put this in bracket

remember the original power there which

is log 4 then plus 4^ log X log X is M

then this is equal to

32 so now from here this power will

multiply together so we have 10 the^ of

M log 4 + 4^ m this is equal to 32 now

let's take note of this a * log B is

equal to log B a^ of a right so here

this will be 10^ of log 4^ m + 4^ m this

is equal to 32 remember this log here

has a base 10 right so take note of this

general rule a log B to the base a is

equal to B so we can see the base of

this logarithm is 10 so we have 10 to

the^ log something to the is 10 so this

will simplify to that something which is

4^ M so 4^ m + 4^ m this is equal to 32

we can see 4^ m is common here so Factor

it out so we are left with 1 + 1 this is

= to 32 from here we have 4^ M * 2 this

is equal to 32 let's go ahead and divide

both sides by

two so we get here 4 to m is = 16 so

here 4^ m is = 16 is 4 2 so from here we

have the same base so we're going to

equate the powers right so from here

this implies that m is equal to

2 now let's recall

back that X is = to 10 to the power of M

so from here we're going to solve for

the value of x easily that X is equal to

10^ of

2 which means that X is = to 100 and

this is the right answer to this problem

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

Interesting Olympiad Math | Equation with logarithmic powers.


hello good day fans welcome back to

another video please subscribe to my

channel if you have not done that please

make sure you like comment and share

here we're given that x - y = 4 x * y =

3 and right over here we're going to

evaluate X Cub - y Cub without first

solving for the value or values of X and

Y in these equations right okay now

we're going to use the formula for the

difference of 2 cu

right so that is x - y cubed + 3 * x * Y

open braet x - y right so from here we

can use this to get the value of x Cub -

y Cub as follows so here we have 4 cubed

right because from the problem here x -

Y is 4 so x - y Cub is 4 cubed then + 3

* X Y that is 3

then * x - y that is 4 right so this is

going to be equal to 4 Cub so that is 64

right and here plus here we have 3 * 3

that is 9 then * 4 that will give us 36

right so from here 64 + 36 so you know

what that's going to give us so that is

equal to 100 and this is the right

answer

thank you for watching please subscribe

to my Channel please like comment and

share bye-bye

An easy Junior Olympiad Math.


good evening and welcome to my YouTube

channel today we have the following

problem with five ways for Negative X is

equal to one all over ten then 25 raised

by S is a quarter Dash our main goal

here is not solving for the value of x

and when we solve for the value of x

Computing the value of 25 raised by S we

need a calculating device here we're

going to use the following indices law 1

over a is equal to a raised by negative

one and again a raised by B then in

black race plus C is the same as a

raised by B times C and again a raise

for B in breaker is plus C is the same

as switching exponents a responsible C

then rest for B okay here we have five

respond Negative X is equal to one over

ten then we have five whole Square

negative one you know

five raised one negative one times x

that is negative one times AC has given

us Negative X is equal to you know one

over ten is ten raised for negative one

okay here you know we have here we can

write this as five raised for X here

then raised by negative 1 is equal to

what 10 raised to power negative one and

right here we can see we have equal we

have equal Powers here so we can equate

the basis so we have five raised for X

is equal to what's 10 okay now upon

squaring both sides of this equation so

we have five raised to x squared is

equal to 10 raised power 2 okay right

here we can rewrite as five raised to

square two then raised for X is equal to

10 raised power 2 is what 100 and right

here we simplify this five raised part 2

here is 25 then raised for X is equal to

100 and this is the right answer to the

problem thank you for viewing and make

sure you subscribe to my YouTube channel

for more

A nice exponential problem.


hello good viewers welcome back once

again let's find a solution to the

following given radical

equation we're given Square OT of 5

+ the cube root of

x this is equal to 3 from here let's

Square both sides of this

equation now this square and this root

sign will cancel out and we have 5 + the

Cub root of x this is equal to 3 2 which

is 9 from here we're going to subtract

five from both sides we get cube root of

x is = to 9 - 5 which is 4 now to get

rid of this cube root sign since we're

solving for x we're going to cube both

sides of this equation so from here we

Cube both sides and here the cube and

the cube root to cancel out and we left

with only X here this is equal to 4

cubed which is 64 and this is the right

answer thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting radical equation | #maths #mathematics


hello welcome back once again today

we're given the following Olympian

mathematics problem we're given A+ B is

= 1 a 2 + b 2 is = 2 so here we'll be

evaluating a cub + B Cubed mind you

three is not the right

answer okay now let's get started from

equation one we have a + b which is

equal to 1 so let's go ahead and square

both sides

the left hand side will become a 2 + b 2

+ 2 * a * B this will be equal to 1 so

we know from equation two this right

over here is equal to two so we have 2 +

2 a b this is equal to 1 subtract two

from both side we get 2 a b is = 1 - 2

which is -1 then divide by

2 then from here we get the product of A

and B which is equal to -1 /

2 now the expression we asked to

evaluate a cubed + B Cubed has the

following ni algebraic identity this is

equal to a + b Cub - 3 * a * B into

braet a + b so this will be equal to We

Know A + B is 1 so we get 1 Cub - 3 * a

b which is -1 / 2 * A + B which is

1 so here we get a cubed + B Cubed is =

to 1 -3 * -1 /2 is pos3 / 2 then POS 3 /

2 * 1 is POS 3 /

2 we know 1 is same as 2 / 2 then we

have + 3 / 2 add the numerators so we

get 5 / 2 so we have therefore a cubed +

B Cub is = 5 / 2 and this is the right

answer thank you for watching please

comment and share bye-bye

Germany | Nice Olympiad Math | Can you solve?


hello welcome back once again today

let's solve this interesting equation

involving square root and cube root

signs so we have here theare root of x

is equal to the cube root of x now let's

go ahead and solve for the values of X

now let's take note of

this the end root of a is equal to a to

the^ of 1 /

n okay so here the left hand side which

isare root of x we're going to write as

X to^ of 1 / 2 then this will be equal

to the cube root of x from the right

hand side we're going to write it as x

to^ 1 / 3 okay now in our next step

we're going to raise both sides of this

equation to a power of a number but what

is this number that we're going to raise

both sides to so how we get this number

is to find the LCM of 2 and 3

so fortunately the LCM of 2 and 3 is

equal to 6 so let's go ahead and raise

both sides of this equation to the power

of 6 so we have X to the^ 1 /

2 6 this will be equal x^ 1 / 3 6 now

let's make use of this property a^ b to^

c this is equal to a^ B * C

so we can see from here 1 2 * 6 will

give us 3 right so we get x to^ 3 this

is equal to 1 / 3 * 6 will give us 2 so

we get here x^ 2 so from here let's

subtract X from both sides to get X Cub

X2 this is equal to 0 from here X2 is

common so we Factor it out so X Cub x²

will give us x - x ID x² will give us 1

then this is equal to Z so when we have

two products which is equal to Z either

the first which is here is equal to 0

which is X2 is equal to 0 or the second

here is equal to 0 which is x -1 is = 0

so here if x is = 0 then X will be squ

otk of 0 which is 0 so we have X is

equal to Z so this is our solution from

here add one to both side we get another

solution that

xal 1 so therefore the given equation

has only two real solutions thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye


hello good day viewers welcome back once

again to my YouTube Channel please if

today is the first day of visiting this

Channel please kindly support the

channel by subscribing please like

comment and share here we're giv that A

+ B is equal to 50 a * B is equal

624 and here we asked to evaluate a 2us

b 2 please this is an Olympian math

problem let us see how to solve this

problem here we record the difference of

two squares a 2 - b

2 is equal to a + b * Aus B right so

this expression right over here is the

same as a + b * Aus B but in the given

problem we have the value for a + b but

we don't have the value for a minus B so

our K here is to find the value for a

minus B okay from equation one we have a

+ B is equal to 50 let us Square both

sides of this equation so here we have a

+ b s is equal to 50 s right so now

recall this property x + y^ 2 is = to x

- Y 2 + 4 x y right so going to use this

property at the left hand side of this

equation so here we have a - b 2 then

plus 4 a b this is equal to 50 2 that is

2,500 right so from the given problem we

have the value for a * b as we can see

it here so we're going to replace this

we have a minus b 2 plus here we have

four * the value for a * B that is

624 and this is equal to

2,500 right so from here we have a - b 2

plus here we have 4 * 64 that will give

us

2496 so 2496 is equal to 2500 right so

from here we'll have Aus b 2 is equal to

2500 -

2496 that is going to give us 4 so from

here we get a - b is equal to you know

plus or minus < TK4 right so this will

give us Plus Plus or - 2

cool so therefore the expression a s - b

2 is going to be equal to a + b MTI a

minus B right so this is equal to A+ B

in the given problem we have it as 2

then 2 * plus or minus 2 right so here

we have plus or minus 2 and this is

going to give us plus or minus 100 and

this is the right answer thank you for

watching please do wa to subscribe to my

YouTube channel

please like comment and share bye-bye

lnteresting Olympiad Math.


hello gooders welcome back once again in

this video we're going to evaluate this

indefinite integral here the integral of

1 /un of 1 - x^2 DX in this video we're

going to practice how to make use of

trigonometric

substitutions now let us talk about this

radical here we have root of 1 - x^2

let's think about this we know 1 - sin s

of theta this will give us cosine squ of

theta so we'll be using this idea to

make a subtitution letting xal sin of

theta since X is squ then s of theta

will be squar so here we let X = sin

Theta now when we differentiate both

side we get DX isal to the derivative of

sin Theta with respect to Theta will

give us cine Theta D Theta right so from

here if you call this integral

I so from here we get I is equal to the

integral 1 over < TK of 1 - x² will be

sin s

thet then time the x is cine Theta D

Theta right so from here we get I is

equal to the integral of 1 all theun of

1 - sin Theta is equal to cosine s Theta

then here cos

Theta D Theta and from here this is

equal to the integral of 1

/un cosine s Theta will give us cosine

Theta then here time cosine Theta D

Theta so this will cancel out from here

remember we're integrating with to Theta

and from here the integral of 1 D Theta

will give us Theta then plus constant

right

okay now what is Theta from the first

substitution here we all know that X is

= to sin Theta if you want to solve for

Theta here then we take the X sign right

or the sign inverse or inverse tangent

of X that is actually what is Theta so

from here we have gotten the value of

theta in terms of X and here we come up

with a solution that the integral is

evaluated as the inverse s of X plus C

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

Indefinite integral | Trigonometric substitution | #maths #mathematics


hello good viewers welcome back once

again today we're going to solve for the

value of x in the following exponential

equation right and one can easily say

that X is equal to 0 here we have

9^ 0 which is same as 9^ 0 which is 1

this equal to which is 1 we can see that

1 is = 1 so X is = z is a

solution but the question here is do we

have all

solution okay now let's go ahead and

check it out right so here we have 9 to^

ofun x this is equal to

x now the idea here is to write the

bases from both sides and index power of

three right because we already know that

9 is 3

2un x this is equal to 27 is as 3^ of 3

X now with this identity a to the^ of b

c this is equal to a to the^ of B * C so

here this will be 3^ 2 * root X which is

2un X then this is equal to 3 to^ 3 * X

here which is 3 x so since we have equal

bases we can equate the powers right so

from here we have that 2 of X that this

implies right okay

2 2un x is = 3 x now let's go ahead and

square both

sides you know why to get rid of this

root sign and again A B C this equal to

a c * b c so here this will become 2 2 *

otk

x² this is equal to here 3 2 * x²

so here this the root sign and the

square will cancel out here we have 2

square which is 4 4 * X here is 4 x this

is equal to 3 2 is 9 so here we have 9

X2 so from the right we have 9 X2 - 4x

this is equal Z so let's go ahead and

factor out X we have 9x - 4 this is

equal Z so from here we have x is = 0

and in the bracket 9x - 4 is also = to Z

so add four to both side we get 9x is =

4 right divide both by 9 we actually

have X is = 4 / 9 and here we have the

solution x is = 0 or X is = 4/ 9 and

these are the solutions thank you for

watching please kindly subscribe to my

YouTube channel also like comment and

share bye-bye right

Interesting exponential equation | #maths #olympiad


hello good day viewers welcome back once

again today we have this sodic equation

and we're going to solve for A and B

we're given that < TK A + B < tk3 this

is equal to 13 All 4 + <

tk3 now for the right hand side of this

equation let us first rationalize it so

here we have

13 all over 4 + < tk3 to rationalize

this we're going to multiply the

numerator and the denominator by the

conjugate form of the denominator so

here we have to multiply by 4 - < tk3

and down here we have to multiply by 4 -

tk3 okay now from the denominator right

over here we can easily see that this is

the difference of two squares remember a

+ b * Aus B this is equal to a 2 - b 2

so this is equal to 13 * 4 - < tk3 then

all over that would be 4 2 - < tk3 2 so

from here we get 13 into bra 4 - < tk3

then all 4 2 is 16 - now this square and

the root sign will cancel out we have

three and from here this is equal to 13

into 4 - < tk3 then all over 16 - 3 that

is 13 now this 13 and this 13 we cancel

out and we left with 4 - root3 so from

here we have the

equation < TK A + B < tk3 this is

actually equal to 4 - < tk3 now from

here let's go ahead and square both

sides to get rid of this root

sign so from here the square and the

root sign we cancel out we'll have a + b

root of 3 this is equal to no when you

expand this we have the first 10 squ

which is 16 plus the 2 10 squ which is 3

- 2 * the first term and the second term

that is 2 * 4 8 * < tk3 that is 8 < tk3

so we have a + b < tk3 this is equal to

here we have that is 19 - 8 < tk3 so

from here we can compare and from here

this implies that a is equal to

19 and B is equal to -8 and this is the

solution thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Interesting surdic equality | #maths #mathematics


good day viewers welcome back to my

Channel today we're having the following

exponential problem please if you have

not subscribed to this Channel Please

Subscribe and turn on the notification

Bell so that you'll be notified anytime

we upload new videos

today we have any following exponential

problem five raised per X all over 7 is

equal to 7 raised to power 5 all over X

how do we solve this problem okay now

recall that the natural logarithm of X

is the same as the logarithm of x to the

base a where e is a last number

okay so I'm going to take the natural

logarithm on both sides of this equation

so I'm going to take the natural

logarithm so I have

not sure logarithm of 5 raised by X all

over 7 is equal to Natural algorithm of

7 raised power 5 1 y x okay remember

that this is one of the properties of

logarithm the natural logarithm of X

raised Y is the same as the power of s

times the entire logarithm okay so here

I'm going to have X all over seven the

natural logarithm of 5 is equal to 5 all

over X the natural logarithm of seven

okay this equation I'm going to rewrite

it as follows so I can write this as X

natural logarithm of 5 all over 7 is

equal to 5 if a natural logarithm of 7

all over X okay so I'm going to cross

multiply so X natural logarithm of 5

times x is equal to x squared natural

logarithm of 5 is equal to

7 times 5 natural logarithm of 7 is 35

natural logarithm of 7 okay now from

here I'm going to divide both sides by

the coefficient of x squared so that is

over natural logarithm of 5. over

natural logarithm of 5 Okay so this and

this get canceled at so we have x

squared

is equal to 35

natural logarithm of seven

all over not sure login number five okay

I'm going to write this

as follows so remember that the natural

logarithm of X all over the natural

logarithm of Y is the same as the

logarithm of x to the base y okay

right so I'm going to write this as

follows

x squared is equal to

times natural logarithm of seven all of

our natural logarithm of 5 and this is

the same as 35

logarithm of 7 to the base 5 okay so I'm

going to take the square root of both

sides so I have square root of x squared

is equal to you know plus or minus

square root of 35 or logarithm of 7 to

the base 5. and we on this way and this

where we get canceled so we are left

with X is equal to you know square root

of

log 7 to the base 5 or

negative square roots of 35

log 7 to the base 5 and this is the

right answer to the problem thank you

for viewing and please make sure you

subscribe to my YouTube channel for more

interesting videos

Solving a nice exponential equation. Not just one value !


hello good viewers welcome back once

again to another video please canly

support this channel by subscribing to

this channel also like comment and share

here we're given the following

interesting exponential equation problem

we're given that 2^ X is = x^ 32 we

asked to find the value of x right okay

now in this video we'll be making use of

the

following identities here we have that a

the^ b into bracket C is equal to c b

right and here A B into bra C isal to a^

B * C okay now let's see how to make use

of these identities in this video we

have that 2 ^ of X is equal to X the^ of

32 right now let's go ahead and square

both sides of this

equation okay at the left hand side

we're going to make use of the first

identity here right so we're going to

have 2^ 2 then of X this is equal to so

we make use of the second identity in

the at the right hand side of this

equation so we're going to have x 32 * 2

which is 64

right okay now from here 2 2 is 4 so we

have 4^ X this is equal to x^ of 64 now

let's go ahead and square both sides

again

so make use of the same identities we

have 4 x this isal to x 64 * 2 which is

128 right so from here we have 16 to the

power of X this is equal to x^ of

128 now let's Square both sides again

wow this is getting interesting

right okay we are looking for something

and we must find it now we do the same

thing 16 s then of X this is equal to x

1 28 * 2 that will give us

right okay now from here 16

squ okay now 16 squ is going to give us

256 right and here we have ra power of X

this is equal to X power of

256 right so from here we can compare

and find the value of x from here right

so we have the bases here x is equal to

256 or we make use of the powers X is

equal to

256 and this is the right answer to the

problem thank you for watching please do

wa to subscribe to my channel also like

comment and share bye-bye

Interesting exponential equation. How to solve for integral value of x ?


hello good viewers welcome back once

again today we're going to learn how to

make use of the logorithm when we have

this kind of exponential

equation we have here 4^ x =

17 here we're going to make use of the

logm from the left hand side we're going

to log it so we have log of four to the^

of x from the right this is equal to log

of 17 right now let us recall this

property of logarithm that the log of a

to the power of B is equal to B multiply

by the log of a right so from here this

x here will come to the front so we have

X MTI by the log of 4 this is equal to

the log of 17 now from here remember

we're solving for x right so we're going

to divide both sides by the coefficient

of x here the coefficient of x is log of

4 right so here we have x * the log of 4

ided by the log of 4 this is equal to

log of 17 / the log of 4 so from here

log of four we cancel log of four right

so here we have the value for x in t of

the logarithm that is the log of 17 /

the log of 4 and when we imput this in

our calculator this is approximately

equal to

2.43 73 and this is the right answer

thank you for watching please kindly

subscribe to my channel also like

comment and share bye-bye

How to solve exponential equation using logarithms | Step by Step | #maths #mathematics


hello welcome back once again today we

have another interesting problem but

this time around we're going to use a

different method and this is called the

deac de poac

formula it's amazing and awesome the

five dict valuation of n Factor by De PO

is given as

Su as n goes from one all the way to

Infinity of the FL function of n / 5 to^

of

n amazing so we're going to find the

number of trading zeros of 2024 Factory

remember in the previous video we used

the legenda formula this around the poly

is an action right so let's go ahead and

find that now so the number of trading

zeros of 224 factoral is going to be

equal to sum as n goes from 1 to

Infinity of 2024 / 5^ n the FL of 8

right if you're not familiar with the

FLW function the FLW function of 6.9

only give us the integral part six six

the Flor function of

3.15 give us the integral part which is

three so that's just the operation of

the L of the FL function where is lumat

coming from am I okay okay let's

continue okay now from here we have the

number of training zeros of 2024

factorial is going to be equal to now

some from here we're going to have Flor

2024 when n = 1 that would be 5 year

plus sum 2024 / 5 2 which is

25 plus here sum oh sum are we still

suming yeah so 2024 / 5 Cub which is one

to5 then plus Flor I mean not some yeah

that is true then 5 to ^ 4 which is 1 25

+ 4 2024 / 5^ of 5 of five that would

give us 3 1 2 5 right then plus and so

on

right one of the most interesting part

of this is that this is going to

converge and the rest of the sum will be

equal to 0 0+ 0 plus 0 plus the way to

Infinity that will still be zero right

so here this divid by this

the flow of it is going to give us 44 so

no need of putting that if you divide

2024 by 5 you're going to have 44

something and the flow of it is just

that 44 which is the integral part then

plus this here the flow of it will give

us 80 + 16 + 3 + 0 + 0 + 0 + 0 dot dot

dot and the sum evaluates to

53 and this was exactly the answer we

got

in our previous video using the legend

formula thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

De - Polignac Method

#maths #physics #olympiad #mathematics #number


hello welcome back once again today

we're going to solve this interesting or

impaired mathematics problem we're given

4^x is = to X so in this video we're

going to find the value for x so let's

get

started we know this

property a^ n is = to 1 all a to the^ n

so we're going to use this at the left

hand side so the left hand side will

become one

All 4 to the^ X this is equal to this

the same as X all 1 so we're going to

cross multiply so x * 4^ X will give us

x * 4 to power x so this is equal to 1 *

1 will give us

1 okay in our one step here let us make

use of this property we know a to the

power of B can be written as e to^ b *

the Ln of a right okay so 4^ X let us

write it so this will become

x e^ x the Ln 4 so this will be equal to

one now our next Target is to make the

power of E equal with the coefficient so

we can see the only difference between

this power and this coefficient is the

natural L of four so let's multiply both

by the natur 4 so this will become X the

natur 4 * e^ X the natur 4 so this is

equal to the natural log of four okay

now we know the natural log of

four is same as the natural log of 2 to

the^ 2 and with the power rule for logm

so this will become two is the power

rule not I don't think so it's just this

property that end of a the power b isal

b the Ln a so this power here is coming

down here so this will come 2 * the Ln

of 2 and we know two here we can write

as e^ the Ln of 2 then M the Ln of 2 so

this is what the right hand side will

become so we get here x * the natur 4 e

* x * natur of 4 so thisal to e^ Ln 2

the Ln 2

so we're going to make use of the Lamb

function so remember that the W of x * e

x is equal to X where the power of E and

the coefficients are equal so when we

use the lambar W function on both sides

of this

equation so this side will simplify to x

* the natural L of 4 so this will be

equal to here we simplify to natural log

of 2

so divide both side by the Natural

Four so this will cancel this so here we

get our solution X is equal to natur of

2 / we know natural of 4 is same as

remember from here Ln of 2 2 which is 2

* Ln of 2 so this will become 2 * Ln of

2 so Ln of 2 we cancel L of 2 so we have

1 / 2

or5 right so this is our solution thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Stanford University 🎓 Entrance Exam | Can you solve ?


hello good day viewers welcome back once

again please kindly subscribe to my

channel also like comment and share here

we're given the following interesting

algebraic problem we have a 2 b 2 plus p

2/ a 2 is =

727 right over here we're going to

evaluate a/ b + B over a

right okay now let's try expanding

the following

binomial A over B + b/ a s right so

we're going to make use of this property

a + b s this is equal to the first term

here squar plus the second term

squ plus 2 * their product right so here

we're going to have this to be equal to

a/ b 2 plus

B / a 2 + 2 * a/ B * B / a right so from

here we have a/ b + b/ a s so this is

equal to you know this is same as a 2/ b

2+ b

2/ a 2 plus you know this we cancel out

so we have two

and remember that a 2 b² + B a² is =

727 right so from here we have a/ b + b/

a 2 this is equal to

727 +

2 so from here we get a/ b + B over

a² this is equal to

729 right so from here we're going to

take root of both sides and from here we

get the value for a/ B plus b/ a which

is equal to plus orus < TK

729 and this is equal to plus or minus

27 and this is the right answer thank

to my channel also like comment and

share bye-bye

Interesting algebraic problem.#maths


good day viewers welcome back to my

Channel today we have the following

interesting question

if a plus b is equal to seven

a raised by three plus b raise power is

equal to 28 then a times B is equal to

Dash Okay so let's see how we're going

to solve this problem now let's recall

the following identity

that a

plus b e raised power 3

minus three times a times B Open Bracket

a plus b

is equal to

e raised power 3 plus b respiratory okay

we're going to use this to solve this

problem okay

now we're going to make substitutions

here

so we already know that a plus b is

equal to seven so we have here seven

respiratory minus no three times a times

B we're looking for a times B then times

a plus b here is 7 is equal to a raise

power 3 plus b raised to power 3 is

already Twenty Eight so you can say in

question

so from here we get seven we raised to

the power 3 which is equal to 3 43 minus

you know three times over here is 21

then times a times B is equal to 28 okay

now we have here 21 a b you know if 343

is crossing over so we have three four

three minus 28 okay so here we have 21 a

b is equal to 343 minus 28 so we have

three one five so from here we divide

both subs equations of AAP so that is

over 21 over 21 and from here we get a

times B to be equal to three one five

divided by 21 which is equal to 15 and

this is the right answer thank you for

everyone and please make sure you

subscribe to my YouTube channel for more

interesting videos

A problem given to a student to solve. a + b = 7, a³ + b³ = 28,  ab = ?  Interesting problem!


hello good viewers welcome back once

again let's solve the following given

exponential equation very quickly we

have here 2 ^ of X cubed this is equal

to 3 now we're going to make use of the

natural logarithm so we're going to Ln

both sides of this equation so here we

have Ln of 2^ of X Cub this is equal to

the Ln of 3 now with the property of

logarithm that the logm of a the power

of B this is equal to B the logarithm of

a right so here the power of two here is

X cubed so this power here is coming

over here so we have X cubed multip by

the Ln of 2 this is equal to the Ln of 3

remember that we're solving for the real

value of x right so here we're going to

divide both side by the Ln of 2 so we

have X cub the Ln of 2 / the Ln of 2

this is equal to the Ln of 3 / the Ln of

2 so from here this we divide out and

here we have X cubed this is equal to

the Ln of 3 / the Ln of 2 now to solve

for x here we're going to take the cube

root of both sides so from here x is

equal to the cube

root of the Ln of 3 / the Ln of 2 and

this is approximately equal to

9314 and this is the right answer thank

you for watching please kindly subscribe

to my channel also like comment and

share bye-bye

Germany | Interesting exponential equation | #maths #mathematics


hello good viewers welcome back once

again here we have X + Y = 2 here x * y

= 3 here we're going to find the value

of x Cub + y

Cub now let's solve this question very

quickly we're going to make use of this

nice algebraic identity that a cubed + B

Cubed = A + B Cubed

then - 3 * a * B into braet a +

b so with the help of this the

expression here we can write it in terms

of x + y and x - Y and that will be

equal to x + y Cub - 3 * x * Y into

braet x + y from here we make the

following substitutions this will be 2

cub

- 3 * x * Y which is 3 * X + Y which is

2 from here this equal 8

minus here 18 which is equal to -10 and

this is the right answer thank you for

watching please kindly subscribe to my

channel also like comment and share

bye-bye

Interesting Junior Olympiad Math.


hello welcome back once again today we

have this nice equation here involving

thetion right from the right hand side

of this equation we have 16 Tetra 3

which is not same as 16 to^ 3 okay for

those of us that are not familiar with

the

tetration so 16 Tetra 3 is same as 16 ra

the power of 16 rais power of 16 in

three places

just like 4 T 2 which is same as 4 to

the^ 4 and not equal to 4 to^ 2 okay so

from the left hand side here we have 2^

2^ X then we'll write the right hand

side as 16 to the^ 16 to^

16 okay so here we have 2^ 2 to^ x so

this will be equal to 16 here which is

the base here is 2 ^ 4 then

16 so from here we have 2^ 2^ X so this

will be equal to 2 remember this

property a^ b^ c isal a^ b * C so this

will be 4 * 16 to the^ of 16 so we can

see we have the same basis from both

sides so we're going to equate the power

so this implies from here that 2 to the^

of X is equal 2 4 is as 2^

2 Then multip by here 16 here is 2^ 4

then to^

16 so from here we have 2^ X is equal to

here we have 2 to the^ of

2 okay so from here we have this multip

by 2 to the power of here will be 4 * 16

which is 64

so here we have 2^ X will be equal to

remember we have the same base so we're

going to add the powers here so we have

2^ 64 + 2 which is 66 so from here we

can see we have the same basis so we're

going to equate the power so this

implies from here that X is equal to 66

and this is the right answer thank you

for watching please kindly subscribe to

my channel also like comment and share

bye-bye

TETRATION PROBLEM | Can you solve ?


hello welcome back once again today we

have this interesting radical equation

and here we're going to solve for the

values of X so let's get

started from the left hand side we have

the root of x plus the root

ofx this is equal to 4 this time let's

take it to the right right so this will

become root

ofx is equal to that would be 4us theun

of X so in our next step let's Square

both sides of this equation so we have

root

of- x^ 2 this will be equal to 4 - otk

x^

2 now here Square will cancel square

root and for the right hand side let us

make use of this interesting identity

for a minus b 2 so this is = to a 2 + b

2 - 2 * a * B so here we only

havex so this will be equal to 4 2 is 16

plus otk X2 is just x - 2 * 4 * otk X

which is 8un

X so 8 root X

right okay so from here

we have to move this term to this side

so remember that this term

here we subtracting 8ot X from this term

so if 16 + xus this will give us this so

This actually mean that 16 + xus this

one will also give us this so that will

become 16 + x

--x so this is equal to 8 < TK X

X so here we have 16 plus you know this

is x - - x which is x + x which is 2x

then this is equal to 8 < TK X so now

let's Square both sides to get rid of

this root sign at the right hand side so

here expanding 16 s is

256 plus 2 x^2 is 4 x 2 + 2 * 16 * 2 x

is 64x X then this is equal to you know

82 is 64 otk x^2 is just X we can see we

have positive 64x from both sides so

subtract 64x from both sides and we are

left with

4x2 plus here 256 is equal to Z 256 is

positive if it's crossing over it to

become negative so we have

4x^2 is = to

-256 divide both sides by

four four we cancel four from here so we

have X2 is equal to -256 / 4 will give

us

-64 take the square root of both sides

so we get X isal to plus or minus the

otk of -64 which is equal to plus or

minus of1 is iota andun of 64 is 8 so

therefore we have our solutions for x x

is equal to plus or minus 8 I which is

an imaginary

solution thank you for watching please

kindly subscribe to my channel also like

comment and share bye-bye

Canada | Nice Olympiad Math problem | Can you solve ?


let's find the derivative of x Cub *

tangent X using the product rule first

let's identify the two functions we're

working with let's call you of x = x

cubed and V of X = tangent X the product

rule states that the derivative of U * v

= u * the derivative of V + V * the

derivative of U let's substitute our

function into the product rule formula

we get X Cub * the derivative of tangent

X Plus tangent x times the derivative of

x cubed now let's find each derivative

separately first the derivative of

tangent X is secant 2 x and the

derivative of x cubed is

3x^2 let's substitute these derivatives

back into our equation we get X Cub *

secant 2 x + Tang x *

3x^2 now we can factor out x^2 from both

terms this gives us x^2 * thean x^ 2 x +

3 tangent X let's review what we did we

use the product rule to break down the

derivative into two parts we found the

derivative of tangent X is secant 2 x

and the derivative of x cubed is

3x^2 then we combine terms and factored

out x^2 to get our final answer and

there we have it the derivative of x Cub

tangent X is x^2 * thean x^ 2 x + 3

tangent X

Finding the Derivative of x^3 tan(x)  Using the Product Rule | Step-by-Step Explanation

In this video, learn to differentiate x^3 tan(x)  using the Product Rule in calculus.

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of x^3 tan(x)
- Formula breakdown and example problem-solving

This example is perfect for calculus beginners or those reviewing the Product Rule!
💡 Don't forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp


let's solve this calculus problem step

by step we need to find the derivative

of the natural logarithm of cine 2 Theta

first let's identify that this is a

composite function where we have the

natural log of another function this

means we'll need to use the chain rule

remember that the derivative of natural

log of U is 1/ U * the derivative of U

now we need to find the derivative of

cine 2 thet this is itself requires the

chain rule since we have a composite

function the derivative of cosine is

negative s and we'll multiply that by

the derivative of what's inside which is

2 Theta the derivative of 2 Theta with

respect to Theta is simply 2 let's

rearrange these terms we can write this

as Sin 2 Theta over cine 2 thet * 2

finally recall that s over cosine is the

definition of tangent therefore our

final answer is -2 * tangent of 2 Theta

let's visualize what this means

geometrically here's the graph of our

original function natural log of cosine

2 Theta in blue and here's its

derivative -2 tangent 2 Theta in red

notice how the derivative is steepest

where our original function changes the

most rapidly and that's how we find and

understand the derivative of natural log

of cosine 2 Theta remember the key steps

were using the chain rule twice and

recognizing the tangent identity at the

end

Understanding the Derivative of [ln(cos2𝜃)]   Step by Step

Question: Find the derivative of [ln(cos2𝜃)]
In this video, we break down the process of finding the derivative of  ln(cos2𝜃)  using the chain rule and trigonometric identities. Join us for a detailed explanation to enhance your calculus skills!
#Calculus #ChainRule #Trigonometry #Derivatives #MathTutorial #hsc #hschighermath #1stpaper
#hschighermath1stpaper
#englishversion #cpmrana


let's solve this differentiation problem

step by step we need to find the

derivative of the product of log base a

of X and natural log of x first let's

recall the product rule the derivative

of a product of functions U * v = u *

the derivative of V + V * the derivative

of U in our case you is log base of X

and V is the natural log of x now let's

apply the product rule we'll have log

base uh of x * the derivative of natural

log of x plus natural log of x times the

derivative of log base a of X let's

recall that the derivative of natural

log of x is 1 /x and the derivative of

log base uh of X can be written as log

base uh of e * 1 /x substituting these

derivatives into our equation we get log

base of x * 1 /x plus natural log of x *

log base a of e * 1X now we can factor

out 1 /x from both terms this gives us 1

/x * the quantity log base a of x +

natural log of x * log base a of e let's

verify this makes sense both terms have

a factor of 1 /x which is what we

typically see when differentiating

logarithmic functions the remaining

terms combine the original logarithmic

expressions in a way that preserves

their relationship through the chain

Rule and that completes our solution

remember when differentiating products

of logarithms always start with the

product rule then use the standard

derivatives of logarithmic functions and

finally combine like terms

Finding the Derivative of {(log_𝑎^𝑥 )(ln𝑥 )} Using the Product Rule |Step-by-Step Explanation.

In this video, learn to differentiate {(log_𝑎^𝑥 )(ln𝑥 )} using the Product Rule in calculus.

What You'll Learn:
- Definition of the Product Rule
- Step-by-step differentiation of {(log_𝑎^𝑥 )(ln𝑥 )}
- Formula breakdown and example problem-solving

This example is perfect for calculus beginners or those reviewing the Product Rule!
💡 Don't forget to LIKE, SHARE, and SUBSCRIBE for more math tutorials!
#Calculus #ProductRule #Differentiation #Derivative #MathTutorial #LearnMath #MathEducation #MathHelp


let's solve this calculus problem step

by step we need to find the derivative

of the natural log of x^2 this is a

perfect opportunity to use the chain

rule the chain rule states that when we

have a composite function we multiply

the derivative of the outer function by

the derivative of the inner function

let's identify our functions the outer

function is the square function anything

squared and the inner function is the

natural logarithm now let's apply the

chain rule First We Take the derivative

of the outer function with respect to U

the derivative of u^2 is 2 U then we

substitute back lnx for you giving us 2

lnx next we need to multiply by the

derivative of the inner function the

derivative of Ln X is 1

/x now we multiply these Parts together

2 Ln x * 1 /x this gives us our final

answer 2 Ln X

/x let's verify this makes sense if we

graph y equal Ln x^2 this is what its

derivative looks like notice how the

derivative is positive for X greater

than one which matches our function's

increasing behavior in that region and

and that's how we find the derivative of

Ln

x^2 using the chain rule

Mastering the Derivative Chain Rule Explained with [(lnx)^2] Example

Question: Find the derivative of [(lnx)^2]
Dive into the world of calculus with this step-by-step tutorial on differentiating logarithmic functions using the chain rule. We’ll break down how to find the derivative of [(lnx)^2]  and understand each part of the process clearly. Perfect for students seeking a deeper understanding of these concepts!
#Calculus #ChainRule #Derivative #MathTutorial #hsc #hschighermath #1stpaper #hschighermath1stpaper #englishversion #cpmrana


step by step we'll use a clever

substitution to simplify our work first

let's look at the substitution we'll use

we'll let x equal tangent of t which

means T equals inverse tangent of X

let's see what happens when we make this

substitution in our original expression

we replace x with Tangent T in the

fraction 2x over 1 + x s now here's

where we use a key trigonometric

identity the expression to tangent T

over 1 + tangent s t is actually equal

to sin of 2T this is a double angle

formula therefore our expression becomes

sin inverse of sin 2T since sin inverse

and S are inverse functions they cancel

out leaving us with Just 2 T now we can

substitute back for T remember that t

equals inverse tangent of X so our

expression becomes 2 * inverse tangent

of X finally we can differentiate this

expression the derivative of inverse

tangent is 1 over 1 + x^2 and we

multiply this by two let's summarize the

key steps first we made a substitution

xal tangent T then we used the double

angle angle formula to simplify the

expression next we recognized that sin

inverse and S cancel out finally we

substituted back and differentiated to

get our answer 2 over 1 + x^2

How to Differentiate Inverse Trigonometric Functions [𝑠𝑖𝑛^(−1)   2𝑥/(1+𝑥^2 )] A Step-by-Step Guide

In this video, we’ll dive into the differentiation of inverse trigonometric functions, using a detailed example involving the derivative of  [𝑠𝑖𝑛^(−1)   2𝑥/(1+𝑥^2 )]  . Follow along as we break it down step-by-step, explaining each part of the process. Perfect for students looking to strengthen their calculus skills!
#Calculus #Differentiation #InverseTrigonometricFunctions #MathTutorial #LearningMath


welcome today we'll find the derivative

of a complex function that combines

exponentials and logarithms our function

is y = a raised to the natural log of

cosine X let's break down the structure

of this function we have an exponential

function with base a and its exponent is

the natural log of cosine X this means

we'll need to use the chain rule

multiple times let's start with our

first application of the chain rule for

an exponential function a to the X its

derivative is a to x * the natural log

of a in our case X is Ln of cine X now

for our second application of the chain

rule we need to find the derivative of

Ln of cine X the derivative of natural

log of U is 1 / U * the derivative of U

we know know that the derivative of cine

X is sinx let's substitute this in

notice that sinx over cine X simplifies

to tangent X this is an important

trigonometric relationship to remember

now let's substitute this back into our

original derivative we get a to the Ln

of cine x * Ln of a Time tangent X but

we can simplify this this further

remember that a to the Ln of something

equals that something so a to the Ln of

cosine X simplifies to just cosine X

therefore our final answer is cosine x *

Ln of a * tangent X or cosine x * Ln of

a Time tangent X let's review what we

learned one we use the chain rule twice

to we apply the derivative rules for

exponentials and logarithms 3 we use the

relationship between s cosine and

tangent 4 and we simplified using

logarithm properties

Finding the Derivative of [a^ln(cos x)] | Advanced Chain Rule Application

QUESTION:
Find the derivative of y = a^(ln(cos x)), where a is a positive constant.

🔍 Master the step-by-step solution to finding the derivative of a^ln(cos x) in this comprehensive calculus tutorial.
In this video, we cover:
✓ Breaking down complex exponential functions
✓ Double application of the chain rule
✓ Working with natural logarithms and trigonometric functions
✓ Simplifying using trigonometric identities
✓ Clear explanations of each mathematical step
Perfect for:
• Calculus I & II students
• AP Calculus preparation
• College math students
• Math competition preparation
📚 Prerequisites:
Basic differentiation rules
Chain rule
Exponential functions
Logarithmic functions
Trigonometric functions
#Calculus #ChainRule #Mathematics #CalcTutorial #MathHelp #APCalculus #MathTutoring


let's find the derivative of cine 2x

using first principles we'll break this

down step by step step one let's start

by defining our function f ofx = cine 2x

then we can write f of x + H = cine of 2

* x + H step two we'll set up the

difference quotient the derivative is

defined as the limit as H approaches Z

of F ofx + hus F ofx all divided by H

step three now comes the clever part

we'll use the cosine difference identity

the cosine of a minus the cosine of b

equals -2 * s of a plus b/ 2 * s of a

minus b/ 2 let's apply this identity to

our problem we'll substitute a = 2x + 2

H and B =

2x now let's simplify what's inside each

s term in the first sign we get 4x + 2H

/ 2 which simplifies to 2x + H in the

second sign we get H for our final steps

we know that the limit as H approaches

zero of s h over H = 1 and when we

substitute H = 0 into s of 2x + H we get

S of 2x let's visualize this result with

a graph here's the original function

cine 2x in blue and here's its

derivative -2 sin 2x in red notice how

at the peaks of cosine 2X the derivative

is zero and at the zeros of cosine 2X

the derivative reaches its maximum or

minimum values and there we have it

we've shown that the derivative of cine

2x = -2 sin 2x both algebraically and

graphically

Understanding the Derivative of cos 2x  Using First Principles

Dive into the fundamentals of calculus as we explore how to determine the derivative of the exponential function cos 2x  using first principles. Follow along with step-by-step explanations and examples to enhance your understanding of derivatives and exponential functions.
#Calculus #Derivatives #ExponentialFunctions #MathTutorial #FirstPrinciples #cos2x


let's solve this complex differentiation

problem step by step we'll differentiate

y equal s^ squar of natural log of

secant x let's organize our solution in

two columns the left column will show

our mathematical steps and the right

column will explain what we're doing at

each step first let's write out what

we're trying to find we want to find the

derivative of sin^2 of natural log of

secant x with respect to X now we'll use

the chain rule the outer function is sin

SAR so we get 2 * sin of natural log of

secant x time the derivative of s of

natural log of secant x next we apply

the chain rule again to the derivative

of s this gives us 2 sin of natural log

of secant x * cosine of natural log of

secant x * the derivative of natural log

of secc x here's where we can use a

useful trigonometric identity 2 * sin *

cosine = s of 2 * the angle also we use

the chain rule for the natural log now

we can substitute the derivative of SEC

X which is secc x * tangent X finally we

can simplify by canceling cant X in the

numerator and denominator giving us our

final answer tangent x * sin of 2 *

natural log of secan x and there we have

it we've successfully differentiated

this complex function using the chain

rule trigonometric identities and

careful step-by-step work

Derivatives: Sin Squared of Log Secant Explained [𝑠𝑖𝑛^2 {ln(sec𝑥)}]

Explore how to differentiate the function  [𝑠𝑖𝑛^2 {ln(sec𝑥)}]  using chain and product rule techniques. This step-by-step breakdown simplifies complex trigonometric and logarithmic derivatives, perfect for students and math enthusiasts. Watch as we unravel each step to provide clarity and insight into calculus.


Welcome to our lesson on logarithmic

differentiation today we'll learn how to

differentiate a complex function using

this powerful technique our challenge is

to find the derivative of y = x * cosine

inverse of X all divided by the square <

TK of 1 - x^2 why do we use logarithmic

differentiation well when we have

products quotients or complex

Expressions taking the natural logarithm

of both sides can simplify our work

significantly by turning products into

sums and quotients into differences

let's solve a step by step step one take

the natural logarithm of both sides we

get Ln of y equals Ln of the entire

right side step two use the properties

of logarithms to break this down the

logarithm of a quotient is the

difference of logarith

and the logarithm of a product is the

sum of

logarithms step three apply these

properties to our equation this gives us

Ln of Y = Ln of x + Ln of cine inverse

of x - 12 Ln of 1 - x^2

step four now we can differentiate both

sides with respect to X remember when we

differentiate Ln of Y we get 1/ y *

dydx this is implicit

differentiation step five let's handle

each term separately the derivative of

Ln of X is 1 /x for Ln of cosine inverse

of X we use the chain Rule and for Ln of

1 - x^2 we get -2X over 1 -

x^2 step six putting it all together we

get this expression now multiply both

sides by y solve for

dydx step seven our final step is to

substitute back the original expression

for y and simplify after combining like

terms we get our final

answer and that's how we use logarithmic

differentiation to find the derivative

of a complex function remember this

technique is especially useful when

dealing with products and quotients

thank you for watching

Mastering Logarithmic Differentiation [(𝑥 〖𝑐𝑜𝑠〗^(−1) 𝑥)/√(1−𝑥^2 )] Complex Calculus Made Simple.

Problem
Differentiate: y =[(𝑥 〖𝑐𝑜𝑠〗^(−1) 𝑥)/√(1−𝑥^2 )]

📚 In this comprehensive calculus tutorial, we break down a challenging derivative problem using logarithmic differentiation. Watch as we solve:
y = (x arccos(x))/√(1-x²)
🎯 What You’ll Learn:
• Why logarithmic differentiation is powerful
• How to break down complex functions
• Essential logarithm properties
• Advanced chain rule applications
• Step-by-step solution methodology
• Common mistakes to avoid
💡 Perfect for:
• Calculus students
• Math majors
• Engineering students
• Test preparation
• Self-study
#Calculus #Mathematics #LogarithmicDifferentiation #MathTutorial #CaluculusTips #MathEducation #EngineeringMath #StudyWithMe #MathHelp #CalculusProblems


let's solve this physics problem about

acceleration a car increases its speed

uniformly from 5 m/s to 45 m/s over a

time period of 10

seconds we need to find the car's

acceleration let's visualize this we

have a car starting at an initial

velocity of 5

m/s and reaching a final velocity of 45

m/s after 10

seconds let's let's list out what we

know the initial velocity U is 5

m/s the final velocity V is 45

m/s and the time T is 10

seconds to find the acceleration we'll

use the

formula acceleration equals change in

velocity divided by time or written

mathematically equals vus U /

T let's substitute our values into the

formula plugging in 45 - 5 m/s in the

numerator and 10 seconds in the

denominator now let's simplify in the

numerator 45 - 5 gives us 40

m/s 40 m/s / 10 seconds gives us 4 m/s

squared therefore the acceleration of

the car is 4 m/s squared this means that

every second the car's velocity

increases by by 4 m/s

Understanding Acceleration: A Step-by-Step Physics Problem

Welcome back to our channel! Today, we’re going to solve a physics problem involving acceleration. We’ll break it down step by step to help you grasp the concept of linear motion under uniform acceleration.
Problem Statement:
A car increases its speed from 5 m/s to 45 m/s  in. We need to find the car’s acceleration.
#Physics #Acceleration #Kinematics #LearnPhysics #ScienceEducation


let's solve this complex differentiation

problem step by step our question is to

differentiate the expression 2 cosecant

2x * cine of natural log of tangent X

let's organize our solution in a two

column format with mathematical steps on

the left and explanations on the right

this will help us understand each step

clearly let's start with our initial

expression

we need to use the product rule here

since we're differentiating a product of

two functions the product rule states

that the derivative of U * v = u Prime *

V + U * V Prime applying the product

rule we get 2 * the sum of cose 2x * the

derivative of cine natural log tangent X

plus cosine natural log tangent x * the

derivative of Co cant 2x now we need to

use the chain rule to find these

derivatives for the first term we get

negative sign of natural log tangent x

times the derivative of natural log

tangent X for the second term we get

Negative cose 2x Cent 2x * the

derivative of 2x let's simplify further

the derivative of natural log tangent X

is 1 / tangent x * SEC 2 x and the

derivative of 2x is simply 2 finally we

can factor out -2 cose 2x from both

terms to get our final answer this gives

us -2 cose 2x * the sum of secant ^ 2 X

over tangent x * sin of natural log

tangent X plus 2 cang 2x * cosine of

natural log t X and there we have it

we've successfully differentiated this

complex expression using the product

rule chain Rule and our knowledge of

trigonometric derivatives remember to

take it step by step when dealing with

complex derivatives like this one

Derivatives Step-by-Step for Advanced Trigonometric Functions [2𝑐𝑜𝑠𝑒𝑐2𝑥 cos {ln(tan𝑥)}]

Unlock the secrets of calculus in this clear and detailed walkthrough of differentiating complex trigonometric functions. In this video, we apply the product and chain rules to break down the differentiation of the expression [2𝑐𝑜𝑠𝑒𝑐2𝑥 cos {ln(tan𝑥)}] . Perfect for calculus students looking to deepen their understanding or anyone needing a quick refresher. Join us for this informative session and boost your calculus skills!
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