# High School Calculus/The Chain Rule
### Chain Rule

When differentiating a square root function or a quantity raised to a power the chain rule is used.
$(ax^{n}+b)^{n}$
Using the chain rule we take the derivative of the entire quantity to the power, and multiply that by the derivative of the interior quantity:
$n(ax^{n}+b)^{n-1}*(anx^{n-1})$
$=an^{2}x^{n-1}(ax^{n}+b)^{n-1}$
What we did was take the power of the quantity and moved it out front. From there we took the derivative of the inside and then multiplied our previous steps by the quantity with n-1 for the power of the quantity.
Ex. 1
$f(x)=(2x+4)^{3}$
In order to differentiate this properly we must use the chain rule. First thing we do is take the power number and derive it.
$3(2x+4)^{2}$
From here we take the derivative of the inside
$3(2)(2x+4)^{2}$
All we have to do now is some minor simplification
$f^{\prime }(x)=6(2x+4)^{2}$
And that is the derivative of $f(x)=(2x+4)^{3}$
Ex. 2
${\sqrt {4x^{5}+x}}$
Remember that a square root is just something raised to the one half power
Ex. 3
$2(x^{3}+x^{2})^{5}$
Ex. 4
$2x(x^{2}+1)^{3}$
Hint: You will have to use the product rule as well as the chain rule
An alternate way of completing a chain rule is considering the problem as a composite function.
In this instance you consider simply taking the derivative of the outside multiplied by the derivative of the inside.
Chain Rule becomes very useful when dealing with a function to a power, or a square root of a function.
The following proof outlines this idea.
PROOF
Let $h(x)=f(g(x)).$
Assume that $x=c$ and that as $g(x)\rightarrow g(c)$ ,that $x\rightarrow c$
$h'(c)=\lim _{x\rightarrow c}{\frac {f(g(x))-f(g(c))}{x-c}}$
$=\lim _{x\rightarrow c}[{\frac {f(g(x))-f(g(c))}{g(x)-g(c)}}*{\frac {g(x)-g(c)}{x-c}}]$ and $g(x)\neq g(c)$
$=[\lim _{x\rightarrow c}{\frac {f(g(x))-f(g(c))}{g(x)-g(c)}}][\lim _{x\rightarrow c}{\frac {g(x)-g(c)}{x-c}}]$
$=f'(g(c))g'(c){\square }.$
In order to find the derivative, you can take that proof into consideration.
You can use any method you deem viable to find the derivative.
An example of this method is as follows:
$h(x)=(3x-1)^{2}.$
Let $g(x)=3x-1$ and $f(x)=x^{2}.$
According to chain rule, $f'(x)=2x.$
In the problem, $g(x)$ is inside of $f(x).$
So, then $f'(g(x))=2g(x).$
But because of the chain rule, we must also multiply by $g'(x).$ Thus giving us, $2g(x)g'(x)$
We also see that, $g'(x)=3.$
So if we put this all together we get, $h'(x)=2(3x-1)(3)$
$h'(x)=6(3x-1).$
You may have noticed that in some cases, a chain rule may be multiplied out to create a simple power rule.
For the previous example, if you multiply the power out, you are left with a simpler equation.
In some cases, chain rule is easier.
$h(x)=(3x-1)^{2}$
$h(x)=(3x-1)(3x-1)$
$h(x)=9x^{2}-6x+1$
$h'(x)=18x-6$
$h'(x)=6(3x-1).$
Another reason the chain rule becomes very effective is when you are dealing with trigonometric functions.
For example, the derivative of $f(x)=sin(2x)$ is not $f'(x)=cos(2x)$.
Instead, by applying chain rule,$f'(x)=2cos(2x).$
This shows that chain rule is useful whenever a trigonometric function is applied to a function other than x.
Sometimes, you may need to apply chain rule more than once.
Here is an example:
$f(x)=sin^{3}(3x)$
$f'(x)=3sin^{2}(3x)cos(3x)(3)$
$f'(x)=9sin^{2}(3x)cos(3x).$
As you can see, first chain rule was applied to find the derivative of the outside, which was the power on the sine function.
Next, the derivative of the sine function itself.
The last part was finding the derivative of the inside of the sine function.
This method required the chain rule twice. Other problems may require more uses of the chain rule.