How To Find $h'(x)$, If $h(x) = F(g(x))$. - Mathematics Stack Exchange

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Learn more about Teams How to find $h'(x)$, if $h(x) = f(g(x))$. Ask Question Asked 12 years, 8 months ago Modified 12 years, 8 months ago Viewed 26k times 2 $\begingroup$

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 $h'(x) = 2x \sqrt{3x^2 + 1}$.

• calculus
• derivatives

Share Cite Follow edited Apr 22, 2012 at 14:46 TMM 10k33 gold badges3636 silver badges5454 bronze badges asked Apr 22, 2012 at 13:59 Cin Sb SangpiCin Sb Sangpi 64366 gold badges2020 silver badges3636 bronze badges $\endgroup$ 2

• 1 $\begingroup$ I think you want to find $fog^{'}(x)$, so you have to use this formula $$fog^{'}(x)=f^{'}(g(x).g^{'}(x)$$. $\endgroup$ – Kns Commented Apr 22, 2012 at 14:09
• $\begingroup$ @Sb Sangpi ,rule for derivate composite function is following $(f(g(x))'=f'(g(x))*g'(x)$ $\endgroup$ – dato datuashvili Commented Apr 22, 2012 at 14:18

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3 Answers 3

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 2 $\begingroup$

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:

$$ \bigl[\,f\bigl(g(x)\bigr)\,\bigr]' =\color{maroon}{ f'\bigl( g(x) \bigr)} \cdot \color{darkgreen}{g'(x) }= \color{maroon}{\sqrt{3x^2+1 }} \cdot\color{darkgreen}{2x}=2x\sqrt{3x^2+1}. $$

Share Cite Follow edited Apr 22, 2012 at 14:16 answered Apr 22, 2012 at 14:09 David MitraDavid Mitra 75.6k1010 gold badges145145 silver badges203203 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$

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: $$(g(f(x)))'=g'(f(x))\cdot f'(x)=\cos(f(x))\cdot 2x=\cos(x^2)\cdot 2x$$

Share Cite Follow edited Apr 22, 2012 at 14:48 TMM 10k33 gold badges3636 silver badges5454 bronze badges answered Apr 22, 2012 at 14:03 chemengchemeng 1,8791313 silver badges1212 bronze badges $\endgroup$ 2

• $\begingroup$ if f'(x) is given,then we are you trying to find f'(x)? $\endgroup$ – dato datuashvili Commented Apr 22, 2012 at 14:06
• $\begingroup$ The derivative of the normal function f'(x) is given. Sb Sangpi is trying to find the derivative of a composite function, that has the form f(g(x)) $\endgroup$ – chemeng Commented Apr 22, 2012 at 14:09

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$h'(x)=f'(g(x))*g'(x)$

$g'(x)=2x$ 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.

Share Cite Follow edited Apr 22, 2012 at 14:41 Joe 4,77766 gold badges3535 silver badges5555 bronze badges answered Apr 22, 2012 at 14:14 dato datuashvilidato datuashvili 9,3361818 gold badges9292 silver badges151151 bronze badges $\endgroup$ Add a comment |

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