Finding derivative with the given information

**youngb11** · February 20th 2011, 09:35 AM

$\displaystyle y=f(\sqrt{x^2+9})$ and $\displaystyle f'(5)=-2$ , find $\displaystyle\frac{dy}{dx}$ when $\displaystyle x=4$

How would I start this question? Greatly appreciate the help!

**FernandoRevilla** · February 20th 2011, 09:53 AM

Use the Chain's Rule to $y=f(\sqrt{x^2+9})$ and substitute $x=4$ .

Fernando Revilla

**youngb11** · February 20th 2011, 10:05 AM

Originally Posted by FernandoRevilla

Use the Chain's Rule to $y=f(\sqrt{x^2+9})$ and substitute $x=4$ .

Fernando Revilla

Substitute $\displaystyle x=4$ in the derivative of $\displaystyle\sqrt{x^2+9}$ ?

**youngb11** · February 20th 2011, 10:36 AM

Answer in the back of the book is $\displaystyle -\frac{8}{5}$ . I'm not sure where to begin.

**FernandoRevilla** · February 20th 2011, 10:39 AM

$\dfrac{dy}{dx}=f'(\sqrt{x^2+9})\cdot \dfrac{x}{\sqrt{x^2+9}}$

Substitute $x=4$ .

Fernando Revilla

**youngb11** · February 20th 2011, 10:45 AM

Originally Posted by FernandoRevilla

$\dfrac{dy}{dx}=f'(\sqrt{x^2+9})\cdot \dfrac{x}{\sqrt{x^2+9}}$

Substitute $x=4$ .

Fernando Revilla

I'm not sure if I understand how $f'(\sqrt{x^2+9})$ works.

Is $f(x)=\sqr{x^2+9}$ ? If so, where do we use the $f'(5)=-2$ information?

**FernandoRevilla** · February 20th 2011, 10:53 AM

Originally Posted by youngb11

I'm not sure if I understand how f'(\sqrt{x^2+9}) works.

$\dfrac{dy}{dx}(4)=f'(\sqrt {4^2+9})\cdot \dfrac{4}{\sqrt {4^2+9}}=f'(5)\cdot \dfrac{4}{5}=(-2)\cdot \dfrac {4}{5}=-\dfrac{8}{5}$

Fernando Revilla

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