Let $f'(x) = \sqrt{3x + 4}$ and $g(x)=x^2-1$. Find $h'(x)$, if $h(x) = f(g(x))$.
I know that $g'(x) = 2x$, but I don't know how to do further. The answer is $h'(x) = 2x \sqrt{3x^2 + 1}$.
$\endgroup$ 23 Answers
$\begingroup$You should be asking "find $h'(x)$ if $h(x)=f\bigl(g(x)\bigr)$".
Use the chain rule: $$\bigl[\,f\bigl(g(x)\bigr)\,\bigr]' = \color{maroon}{f'\bigl(g(x)\bigr)}\cdot\color{darkgreen}{ g'(x)}.$$
We aren't told what $f$ is, but we do know that $$f'(\color{darkblue}{x})=\sqrt{\,3\color{darkblue}{x}+4}.$$ Then
$$\color{maroon}{f'\bigl(g(x)\bigr)}=f'(\color{darkblue}{x^2-1})=\sqrt{ 3(\color{darkblue}{x^2-1})+4}=\color{maroon}{\sqrt{3x^2+1}}.$$
Also $\color{darkgreen}{g'(x)}=(x^2-1)'=\color{darkgreen}{2x}$.
So, using the chain rule:
$$ \bigl[\,f\bigl(g(x)\bigr)\,\bigr]' =\color{maroon}{ f'\bigl( g(x) \bigr)} \cdot \color{darkgreen}{g'(x) }= \color{maroon}{\sqrt{3x^2+1 }} \cdot\color{darkgreen}{2x}=2x\sqrt{3x^2+1}. $$
$\endgroup$ $\begingroup$Hint:
The general rule for calculating the derivative of a composite functions is: $$(g(f(x)))'=g'(f(x))\cdot f'(x)$$
For example, let $f(x)=x^2$ and $g(x)=\sin(x)$. Then $f'(x)=2x$ and $g'(x)=\cos(x)$. For the composite function $g(f(x))$ you get: $$(g(f(x)))'=g'(f(x))\cdot f'(x)=\cos(f(x))\cdot 2x=\cos(x^2)\cdot 2x$$
$\endgroup$ 2 $\begingroup$$h'(x)=f'(g(x))*g'(x)$
$g'(x)=2x$ as you said
Now what is $f'(g(x))$ equal to? $f'(g(x))=\sqrt{(3(x^2-1)+4)}$ which is equal to $\sqrt{3x^2+1}$. Now if you multiply, you will get the desired result.
$\endgroup$
