datavorous/entrance-exam-dataset huggingface dataset:
What is the definition of the term “Primary Structure”?

Options:
A Primary Structure refers to the first level of protein structure where amino acids are arranged in a specific sequence.
B Primary Structure refers to the overall three-dimensional shape of a protein.
C Primary Structure refers to the interactions between different protein subunits.
D Primary Structure refers to the secondary structure elements, such as alpha helices and beta sheets, in a protein.

The correct answer is:
Primary Structure refers to the first level of protein structure where amino acids are arranged in a specific sequence.

Solution:
Correct Option is : (A)
Primary Structure refers to the first level of protein structure where amino acids are arranged in a specific sequence.

Topic: Chemistry, Biomolecules, NEET
What is the definition of twisted aestivation?

Options:
A When the margins of sepals or petals overlap one another but not in any particular direction
B When sepals or petals overlap one another in a particular direction
C When sepals or petals just touch one another at the margin, without overlapping
D When one margin of the appendage overlaps that of the next one and so on

The correct answer is:
When one margin of the appendage overlaps that of the next one and so on

Solution:
Twisted aestivation refers to a situation where one margin of the sepals or petals overlaps that of the next one, and so on.

Topic: Biology, Morphology of Flowering Plants, NEET
What is the degree of the differential equation
$\left(1+\frac{d y}{d x}\right)^{4}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2} ?$

Options:
A 1
B 2
C 4
D 8

The correct answer is:
2

Solution:
The given differential equation is
$\left(1+\frac{d y}{d x}\right)^{4}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$
From above it is clear that degree of given differential equation is 2 .
Because degree is the power of highest order derivative.

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 2)
What is the degree of the differential equation
$\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3 / 2}=\mathrm{k} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}} ? \quad$

Options:
A 4
B 3
C 2
D 1

The correct answer is:
2

Solution:
The given differential equation is
$$
\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3 / 2}=\mathrm{k} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}
$$
To express it as a polynomial of derivatives we square both side,
$$
\Rightarrow\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\mathrm{k}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right)^{2}
$$
Highest derivative has power $=2$ Degree of differential equation $=2$.

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 2)
What is the degree of the differential equation
$$
\frac{d^{2} y}{d x^{2}}-\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}=0 ?
$$

Options:
A 1
B 2
C 3
D 6

The correct answer is:
2

Solution:
Given differential equation is $\frac{d^{2} y}{d x^{2}}-\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}$
On squaring both the sides,
$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=1+\left(\frac{d y}{d x}\right)^{3}$
Since, degree of the differential equation is the power of highest order derivative. Therefore from above it is clear that degree of equation is 2 .

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 1)
What is the degree of the differential equation
$\left(\frac{d^{3} y}{d x^{3}}\right)^{2 / 3}+4-3\left(\frac{d^{2} y}{d x^{2}}\right)+5\left(\frac{d y}{d x}\right)=0 ? \quad$

Options:
A 3
B 2
C $2 / 3$
D Not defined

The correct answer is:
2

Solution:
Degree of a differential equation is the power to which the highest derivative is raised when it is expressed as polynomial of derivatives. Given equation is
$\left(\frac{d^{3} y}{d x^{3}}\right)^{2 / 3}-3\left(\frac{d^{2} y}{d x^{2}}\right)+5\left(\frac{d y}{d x}\right)+4=0$
$\Rightarrow\left(\frac{d^{3} y}{d x^{3}}\right)^{2 / 3}=3 \frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}-4$
Cube on both side,
$\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\left[3 \frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}-4\right]^{3}$
Hence, degree $=2$

Topic: Mathematics, Differential Equations, NDA, NDA 2011 (Phase 2)
What is the degree of the differential equation
$\left(\frac{d^{3} y}{d x^{3}}\right)^{3 / 2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2} ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Consider the given differential equation,
$\left(\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}\right)^{\frac{3}{2}}=\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}$
In order to find degree, differential equation should be free from fractional indices.
Now, squaring eqn (i) both sides, we get
$\left(\frac{d^{3} y}{d x^{3}}\right)^{3}=\left(\frac{d^{2} y}{d x^{2}}\right)^{4}$
Since, the power of highest order derivative is 3 , therefore degree $=3$.

Topic: Mathematics, Differential Equations, NDA, NDA 2014 (Phase 2)
What is the degree of the differential equation
$\left(\frac{d^{4} y}{d x^{4}}\right)^{3 / 5}-5 \frac{d^{3} y}{d x^{3}}+6 \frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}+5=0 ?$

Options:
A 5
B 4
C 3
D 2

The correct answer is:
3

Solution:
Consider differential equation
$\left(\frac{d^{4} y}{d x^{4}}\right)^{\frac{3}{5}}-5 \frac{d^{3} y}{d x^{3}}+6 \frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}+5=0$
$\Rightarrow\left(\frac{d^{4} y}{d x^{4}}\right)^{\frac{3}{5}}=5 \frac{d^{3} y}{d x^{3}}-6 \frac{d^{2} y}{d x^{2}}+8 \frac{d y}{d x}-5$
$\Rightarrow\left(\frac{\mathrm{d}^{4} \mathrm{y}}{\mathrm{d} \mathrm{x}^{4}}\right)^{3}=\left(5 \frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{d} \mathrm{x}^{3}}-6 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+8 \frac{\mathrm{dy}}{\mathrm{dx}}-5\right)^{5}$
So, highest order derivative $=4$, degree $=3$

Topic: Mathematics, Differential Equations, NDA, NDA 2013 (Phase 1)
What is the degree of the differential equation $\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+2\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}-\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0 ?$

Options:
A 6
B 3
C 2
D 1

The correct answer is:
1

Solution:
Given differential equation
$$
\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+2\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)-\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0
$$
Since exponent of highest order derivative is 1 therefore degree $=1$

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 2)
What is the degree of the differential equation
$\mathrm{k} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{3}\right]^{3 / 2}$, where $\mathrm{k}$ is a constant?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
In the given equation,
$\mathrm{K} \cdot \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{3}\right]^{3 / 2}$
Squaring both the sides,
$\mathrm{K}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{3}\right]^{3}$
Degree of a differential equation is the highest power of the highest derivative in equation when derivatives are expressed as polynomial. Here degree of differential equation is $2 .$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 1)
What is the degree of the differential equation
$\mathrm{y}=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-1} ?$

Options:
A 1
B 2
C $-1$
D Degree does not exist.

The correct answer is:
2

Solution:
Given differential equation is
$$
\mathrm{y}=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-1}
$$
Multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$
$$
\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+1
$$
Since power of highest order derivative is 2 . $\therefore$ degree $=2$

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 1)
What is the degree of the equation
$\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]=\left[\mathrm{y}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{\frac{1}{4}} ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
The given differential equation is:
$$
\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[\mathrm{y}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{1 / 4}
$$
This can be re-written as by squaring both the sides to the power 4 to make it a polynomial of derivative.
$$
\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]^{4}=\left[\mathrm{y}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]
$$
Power of highest derivatives 4 , So, degree of the equation is 4 .

Topic: Mathematics, Differential Equations, NDA, NDA 2006 (Phase 1)
What is the degree of the following differential equation?
$\left(\frac{d^{3} y}{d x^{3}}\right)^{2 / 3}+4-3 \frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}=0$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
The given differential equation can be rewritten as
$$
\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\left(3 \frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}-4\right)^{3}
$$
$\therefore$ Degree of differential equation is $2 .$ (\because Degree is the power of the highest order derivative)

Topic: Mathematics, Differential Equations, NDA, NDA 2011 (Phase 1)
What is the derivative of $\sqrt{\frac{1+\cos x}{1-\cos x}}$ ?

Options:
A $\frac{1}{2} \sec ^{2} \frac{x}{2}$
B $-\frac{1}{2} \operatorname{cosec}^{2} \frac{x}{2}$
C $-\operatorname{cosec}^{2} \frac{x}{2}$
D None of these

The correct answer is:
$-\frac{1}{2} \operatorname{cosec}^{2} \frac{x}{2}$

Solution:
Let $\mathrm{y}=\sqrt{\frac{1+\cos \mathrm{x}}{1-\cos \mathrm{x}}}=\frac{\sqrt{2} \cos \frac{\mathrm{x}}{2}}{\sqrt{2} \sin \frac{\mathrm{x}}{2}}=\cot \frac{\mathrm{x}}{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=-\operatorname{cosec}^{2} \frac{\mathrm{x}}{2} \cdot \frac{1}{2}=-\frac{1}{2} \operatorname{cosec}^{2} \frac{\mathrm{x}}{2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2014 (Phase 1)
What is the derivative of $\sqrt{\frac{1+\cos x}{1-\cos x}}$ ?

Options:
A $\frac{1}{2} \sec ^{2} \frac{x}{2}$
B $-\frac{1}{2} \operatorname{cosec}^{2} \frac{x}{2}$
C $-\operatorname{cosec}^{2} \frac{x}{2}$
D None of these

The correct answer is:
$-\frac{1}{2} \operatorname{cosec}^{2} \frac{x}{2}$

Solution:
Let $\mathrm{y}=\sqrt{\frac{1+\cos \mathrm{x}}{1-\cos \mathrm{x}}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=-\operatorname{cosec}^{2} \frac{\mathrm{x}}{2} \cdot \frac{1}{2}=-\frac{1}{2} \operatorname{cosec}^{2} \frac{\mathrm{x}}{2}$

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 1)
What is the derivative of $\cos ^{-1}\left(\frac{2 \cos x+3 \sin x}{\sqrt{13}}\right)$?

Options:
A $\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$
B $-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$
C 0
D 1

The correct answer is:
1

Solution:
The given function $f(x)=\cos ^{-1}\left(\frac{2 \cos x+3 \sin x}{\sqrt{13}}\right\}$
can be written as
$\cos ^{-1}\left(\cos x \cdot \frac{2}{\sqrt{13}}+\sin x \cdot \frac{3}{\sqrt{13}}\right\}$ ...(i)
Let $\frac{2}{\sqrt{13}}=\cos \theta$ and $\frac{3}{\sqrt{13}}=\sin \theta$
$\mathrm{So},(\mathrm{i}), \cos ^{-1}(\cos \mathrm{x} \cos \theta+\sin \mathrm{x} \sin \theta)$
where $\theta=\tan ^{-1}\left(\frac{3}{2}\right)$
$=\cos ^{-1}(\cos (x-\theta))=x-\theta$
hence, $f^{\prime}(x)=1 \quad(\because \theta$ is a constant $)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 2)
What is the derivative of $f(x)=\sqrt{1-x^{2}}$ with respect to $\mathrm{g}(\mathrm{x})=\sin ^{-1} \mathrm{x}$, where $|\mathrm{x}| \neq 1 ?$

Options:
A $\mathrm{x}$
B $-\mathrm{x}$
C $\frac{x}{1-x^{2}}$
D $-\frac{x}{1-x^{2}}$

The correct answer is:
$-\mathrm{x}$

Solution:
Here, $\frac{\mathrm{d}\left(\sqrt{1-\mathrm{x}^{2}}\right)}{\mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)}=\frac{\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1-\mathrm{x}^{2}}\right)}{\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)}$ is to be found.
$f(x)=\sqrt{1-x^{2}}$
So, $f^{\prime}(x)=\frac{d}{d x}\left(\sqrt{1-x^{2}}\right)=f^{\prime}(x) \frac{1}{2 \sqrt{1-x^{2}}}(-2 x)$
$\Rightarrow f^{\prime}(x)=\frac{-x}{\sqrt{1-x^{2}}}$ ...(i)
Also, $g(x)=\sin ^{-1} x$
$\mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$ ...(ii)
$\therefore \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}=\frac{-\mathrm{x} / \sqrt{1-\mathrm{x}^{2}}}{1 / \sqrt{1-\mathrm{x}^{2}}}$ $\quad$ (using Eqs. (i) and (ii))
$=-\mathrm{x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 2)
What is the derivative of $f(x)=\frac{7 x}{(2 x-1)(x+3)}$?

Options:
A $-\frac{3}{(x+3)^{2}}-\frac{2}{(2 x-1)^{2}}$
B $-\frac{3}{(x+3)^{2}}-\frac{1}{(2 x-1)^{2}}$
C $\frac{3}{(x+3)^{2}}+\frac{1}{(2 x-1)^{2}}$
D $\frac{3}{(x+3)^{2}}+\frac{2}{(2 x-1)^{2}}$

The correct answer is:
$-\frac{3}{(x+3)^{2}}-\frac{2}{(2 x-1)^{2}}$

Solution:
Given function is $f(x)=\frac{7 x}{(2 x-1)(x+3)}$
Breaking into partial fraction
We get, $f(x)=\frac{1}{2 x-1}+\frac{3}{x+3}$
Differentiating w.r.t. $\mathrm{x}$, we get
$f^{\prime}(x)=-\frac{2}{(2 x-1)^{2}}-\frac{3}{(x+3)^{2}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 2)
What is the derivative of $\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}|?$

Options:
A $|\mathrm{x}|+\mathrm{x}$
B $2 \mathrm{x}$
C $2|x|$
D $-2|x|$

The correct answer is:
$2|x|$

Solution:
Given that $f(x)=x|x|$
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \cdot \frac{|\mathrm{x}|}{\mathrm{x}}+|\mathrm{x}|\left[\right.$ Since $\left.\mathrm{f}^{\prime}(\mathrm{uv})=\mathrm{u} . \mathrm{f}^{\prime}(\mathrm{v})+\mathrm{v} . \mathrm{f}^{\prime}(\mathrm{u})\right]$
$=|\mathrm{x}|+|\mathrm{x}|=2|\mathrm{x}|$
Here, $\mathrm{u}=\mathrm{x}, \mathrm{v}=|\mathrm{x}|$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
What is the derivative of $\log _{10}\left(5 \mathrm{x}^{2}+3\right)$ with respect to $\mathrm{x}$?

Options:
A $\frac{x \log _{10} e}{5 x^{2}+3}$
B $\frac{2 \mathrm{x} \log _{10} \mathrm{e}}{5 \mathrm{x}^{2}+3}$
C $\frac{10 \mathrm{x} \log _{10} \mathrm{e}}{5 \mathrm{x}^{2}+3}$
D $\frac{10 \times \log _{\mathrm{e}} 10}{5 \mathrm{x}^{2}+3}$

The correct answer is:
$\frac{10 \mathrm{x} \log _{10} \mathrm{e}}{5 \mathrm{x}^{2}+3}$

Solution:
$y=\log _{10}\left(5 x^{2}+3\right)$
$\frac{d y}{d x}=\frac{d}{d x}\left(\log _{10}\left(5 x^{2}+3\right)\right)=\frac{1}{5 x^{2}+3} \times \log _{10} e \times 10 x$
$=\frac{10 x \log _{10} e}{5 x^{2}+3}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 1)
What is the derivative of
$\left(\log _{\tan x} \cot x\right)\left(\log _{\cot x} \tan x\right)^{-1}$ at $x=\frac{\pi}{4} ?$

Options:
A $-1$
B ${0}$
C 1
D $\frac{1}{2}$

The correct answer is:
${0}$

Solution:
The given function, $f(x)=\left(\log _{\tan x} \cot x\right)\left(\log _{\cot x} \tan x\right)^{-1}$
$=\left(\frac{\log \cot x}{\log \tan x}\right)\left(\frac{\log \tan x}{\log \cot x}\right)^{-1}$
$=\left(\frac{\log \cot x}{\log \tan x}\right)\left(\frac{\log \cot x}{\log \tan x}\right)$
$=\left(\frac{\log \cot x}{\log \tan x}\right)^{2}=\left(\frac{\log \frac{1}{\tan x}}{\log \tan x}\right)^{2}=\left(-\frac{\log \tan x}{\log \tan x}\right)^{2}=1$
$\Rightarrow \mathrm{f}(\mathrm{x})=1 \quad$ (constant function).
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0$
and this is true for $0 < x < \frac{\pi}{2}$
$\Rightarrow f^{\prime}\left(\frac{\pi}{4}\right)=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 2)
What is the derivative of $\log _{x} 5$ with respect to $\log _{5} x$?

Options:
A $-\left(\log _{5} x\right)^{-2}$
B $\left(\log _{5} x\right)^{-2}$
C $-\left(\log _{x} 5\right)^{-2}$
D $\left(\log _{x} 5\right)^{-2}$

The correct answer is:
$-\left(\log _{5} x\right)^{-2}$

Solution:
Let $u_{1}=\log _{x} 5$ and $u_{2}=\log _{5} x$
$\Rightarrow u_{1}=\frac{\log _{e} 5}{\log _{e} x}$ and $u_{2}=\frac{\log _{e} x}{\log _{e} 5}$
On differentiating w.r.t. $x$, we get
$\frac{d u_{1}}{d x}=\left[\frac{\log _{e} x(0)-\left(\frac{1}{x}\right)}{\left(\log _{e} x\right)^{2}}\right] \log _{e} 5=-\frac{\log _{e} 5}{x\left(\log _{e} x\right)^{2}}$
and $\frac{d u_{2}}{d x}=\frac{1}{x \log _{e} 5}$
$\therefore \frac{d u_{1}}{d u_{2}}=\frac{d u_{1} / d x}{d u_{2} / d x}=-\frac{\log _{e} 5}{x\left(\log _{e} x\right)^{2}} \times x \log _{e} 5$
$=-\left(\frac{\log _{e} 5}{\log _{e} x}\right)^{2}=-\left(\log _{x} 5\right)^{2}=-\left(\log _{5} x\right)^{-2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
What is the derivative of $\sec ^{2}\left(\tan ^{-1} x\right)$ with respect to $x$ ?

Options:
A $2 \mathrm{x}$
B $x^{2}+1$
C $\mathrm{x}+1$
D $x^{2}$

The correct answer is:
$2 \mathrm{x}$

Solution:
Let $\mathrm{y}=\sec ^{2}\left(\tan ^{-1} \mathrm{x}\right)$
Let $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta$
$y=\sec ^{2} \theta=1+\tan ^{2} \theta=1+x^{2}$
$\therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(1+x^{2}\right)=2 x$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2019 (Phase 1)
What is the derivative of $\sin ^{2} x$ with respect to $\cos ^{2} x ?$

Options:
A $\tan ^{2} x$
B $\cot ^{2} x$
C $-1$
D 1

The correct answer is:
$-1$

Solution:
Let $u=\sin ^{2} x$ and $v=\cos ^{2} x$
$\Rightarrow \quad \frac{d u}{d x}=2 \sin x \cos x=\sin 2 x$
and $\frac{d v}{d x}=-2 \sin x \cos x=-\sin 2 x$
$\therefore \quad \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{-\sin 2 x}{\sin 2 x}=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
What is the derivative of $\sin (\sin x)$?

Options:
A $\cos (\cos x)$
B $\cos (\sin x)$
C $\cos (\sin x) \cos x$
D $\cos (\cos x) \cos x$

The correct answer is:
$\cos (\sin x) \cos x$

Solution:
$\frac{\mathrm{d}}{\mathrm{dx}} \sin (\sin \mathrm{x})=\cos (\sin \mathrm{x}) \cdot \cos \mathrm{x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 2)
What is the derivative of $\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3 / 2}}\right)$ at $x=1 ?$

Options:
A $-\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{3}{2}$
D $1$

The correct answer is:
$-\frac{1}{4}$

Solution:
Let $y=\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3 / 2}}\right)=\tan ^{-1} \frac{\sqrt{x}-x}{1+\sqrt{x}, x}$
$=\tan ^{-1} \sqrt{\mathrm{x}}-\tan ^{-1} \mathrm{x}$
On differentiating w.r.t.x, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}} \cdot \frac{1}{2 \sqrt{\mathrm{x}}}-\frac{1}{1+\mathrm{x}^{2}}$
Now, $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=1}=\frac{1}{1+1} \cdot \frac{1}{2}-\frac{1}{1+1}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
What is the derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ with respect to $\tan ^{-1} \mathrm{x} ?$

Options:
A 0
B $\frac{1}{2}$
C 1
D $\mathrm{x}$

The correct answer is:
$\frac{1}{2}$

Solution:
Let $\mathrm{y}=\tan ^{-1}\left[\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right]$ and $\mathrm{u}=\tan ^{-1} \mathrm{x}$
Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$
Then, $y=\tan ^{-1}\left[\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right]$
$=\tan ^{-1}\left[\frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\right]$
$=\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right]=\tan ^{-1}\left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right]$
$=\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right]=\tan ^{-1}\left[\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2}, \cos \frac{\theta}{2}}\right]$
$\left(\begin{array}{r}\because 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2} \text { and } \\ \sin x=2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\end{array}\right)$
$=\tan ^{-1}\left[\tan \frac{\theta}{2}\right]$
$\Rightarrow y=\frac{\theta}{2} \Rightarrow y=\frac{\tan ^{-1} x}{2} \quad\left[\because \theta=\tan ^{-1} x\right]$
$\Rightarrow y=\frac{u}{2}$
$\frac{\mathrm{dy}}{\mathrm{du}}=\frac{1}{2}$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 1)
What is the derivative of the function $\quad[2018-I]$
$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\tan \mathrm{x}}+\ln (\sec \mathrm{x})-\mathrm{e}^{\ell \mathrm{n} \mathrm{x}}$ at $\mathrm{x}=\frac{\pi}{4} ?$

Options:
A $\frac{\mathrm{e}}{2}$
B e
C $2 \mathrm{e}$
D $4 \mathrm{e}$

The correct answer is:
$2 \mathrm{e}$

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\tan \mathrm{x}}+\ln (\sec \mathrm{x})-\mathrm{e}^{\mathrm{fn} \mathrm{x}}$
$\begin{array}{l}f^{\prime}(\mathrm{x})=\mathrm{e}^{\tan \mathrm{x}} \sec ^{2} \mathrm{x}+\frac{1}{\sec \mathrm{x}} \cdot \sec x \tan \mathrm{x}-1 \\ \left(\because \mathrm{e}^{\ln \mathrm{x}}=\mathrm{x} \text { and } \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1\right)\end{array}$
$f^{\prime}\left(\frac{\pi}{4}\right)=\mathrm{e}^{\tan \frac{\pi}{4}} \cdot \sec ^{2} \frac{\pi}{4}+\tan \frac{\pi}{4}-1$
$=2 \mathrm{e}+1-1$
$=2 \mathrm{e}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 1)
What is the derivative of $|x-1|$ at $x=2$?

Options:
A $-1$
B 0
C 1
D Derivative does not exist

The correct answer is:
1

Solution:
$f(x)=|x-1|$
Redefined the function $\mathrm{f}(\mathrm{x})$
$f(x)=\left\{\begin{array}{ll}1-x, & x < 1 \\ x-1, & x>1\end{array}\right.$
$\mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{ll}-1 ; & \mathrm{x} < 1 \\ 1 ; & \mathrm{x}>1\end{array}\right.$
$\therefore \mathrm{f}^{\prime}(2)=1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 2)
What is the derivative of $x^{3}$ with respect to $x^{2} ?$

Options:
A $3 x^{2}$
B $\frac{3 x}{2}$
C $x$
D $\frac{3}{2}$

The correct answer is:
$\frac{3 x}{2}$

Solution:
$\mathrm{U}=\mathrm{x}^{3}$
$\frac{\mathrm{dU}}{\mathrm{dx}}=3 \mathrm{x}^{2}$ ...(1)
$\mathrm{V}=\mathrm{x}^{2}$
$\frac{\mathrm{dV}}{\mathrm{dx}}=2 \mathrm{x}$ ...(2)
From (1) and (2)
$\frac{\mathrm{dU}}{\mathrm{dV}}=\frac{3 \mathrm{x}^{2}}{2 \mathrm{x}}=\frac{3}{2} \mathrm{x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 2)
What is the derivative of $x \sqrt{a^{2}-x^{2}}+a^{2} \sin ^{-1}\left(\frac{x}{a}\right)$?

Options:
A $\sqrt{a^{2}-x^{2}}$
B $2 \sqrt{a^{2}-x^{2}}$
C $\sqrt{x^{2}-a^{2}}$
D $2 \sqrt{x^{2}-a^{2}}$

The correct answer is:
$\sqrt{a^{2}-x^{2}}$

Solution:
Let $y=x \sqrt{a^{2}-x^{2}}+a^{2} \sin ^{-1}\left(\frac{x}{a}\right)$
Differentiate both side w.r.t ' $x$ ' we get
$\begin{aligned} \frac{d y}{d x} &=x \frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot(-2 x)+\frac{a^{2}}{\sqrt{1-\frac{x^{2}}{a^{2}}}} \cdot \frac{1}{a} \\ &=\frac{-x^{2}}{\sqrt{a^{2}-x^{2}}}+\frac{a^{2} \cdot a}{\sqrt{a^{2}-x^{2}}} \cdot \frac{1}{a} \\ &=\frac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}=\sqrt{a^{2}-x^{2}} \end{aligned}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
What is the derivative of y as a function of $\mathrm{x}$ with respect to $\mathrm{x}$ for $\mathrm{x} < 0 ?$
Consider the equation $\mathrm{x}+|\mathrm{y}|=2 \mathrm{y}$

Options:
A 2
B 1
C $\frac{1}{2}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{3}$

Solution:
$y=\frac{1}{3} x$ for $x < 0$.
Hence $\frac{d y}{d x}=\frac{1}{3}$
Option (d) is correct.

Topic: Mathematics, Differentiation, NDA, NDA 2016 (Phase 1)