# Thread: Finding derivative with the given information

1. ## Finding derivative with the given information

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

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

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

Fernando Revilla

3. 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 $\displaystyle\sqrt{x^2+9}$?

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

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

Substitute $x=4$ .

Fernando Revilla

6. 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?

7. 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