Chain rule problem: given $f(x)=\sqrt{4x+7}$ and $g(x)=e^{x+4}$, compute $f(g(x))'$.

Question

Question:

Given the functions $f(x)=\sqrt{4x+7}$ and $g(x)=e^{x+4}$, compute $f(g(x))'$.

My Approach:

I have found that found that $f(g(x))=\sqrt{4e^{x+4}+7}$. Should I now just differentiate it to get my answer or is there any simpler method to solve this problem. Any helpful suggestions or answers.

Answer 1

We have

$$f(x)=\sqrt{4x+7}$$

$$g(x)=e^{x+4}$$

Note that the composite function $f(g(x))$ is

$$f(g(x))=\sqrt{4e^{x+4}+7}$$

now you can comput directly $[f(g(x))]'$ as a function of $x$ with the ordinary rules or apply the chain rule for composite functions

$$[f(g(x))]'=f'(g(x))\cdot g'(x)$$

the result is obviously the same.

Indeed by ordinary rules

$$[f(g(x))]'=\frac{4e^{x+4}}{2\sqrt{4e^{x+4}+7}} =\frac{2e^{x+4}}{\sqrt{4e^{x+4}+7}}$$

and by chain rule

$$[f(g(x))]'=f'(g(x))\cdot g'(x)=\frac{4}{2\sqrt{4e^{x+4}+7}}\cdot e^{x+4}=\frac{2e^{x+4}}{\sqrt{4e^{x+4}+7}}$$

Answer 2

use the formula $$f(g(x))'=f'(g(x))\cdot g'(x)$$

Answer 3

$f(g(x))'=f'(g(x))\cdot g'(x)=\frac{4}{2\sqrt{4g(x)+7}}\cdot e^{x+4}=\frac{2e^{x+4}}{\sqrt{4e^{x+4}+7}}$

Answer 4

A bit of trickery :

$y:= (4e^{x+4}+7)^{1/2};$

$y^2= 4e^{x+4} +7.$

Differentiating both sides with resp. to x:

$2y\dfrac{dy}{dx} = 4e^{x+4};$

$\dfrac {dy}{dx}= \dfrac {4e^{x+4}}{2y};$

$\dfrac {dy}{dx} = \dfrac {2e^{x+4}}{(4e^{x+4}+7)^{1/2}}.$