Thread: Finding derivative with the given information

1. Finding derivative with the given information

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

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

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

Fernando Revilla

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

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

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

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

Substitute $\displaystyle x=4$ .

Fernando Revilla

6. Originally Posted by FernandoRevilla
$\displaystyle \dfrac{dy}{dx}=f'(\sqrt{x^2+9})\cdot \dfrac{x}{\sqrt{x^2+9}}$

Substitute $\displaystyle x=4$ .

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

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

7. Originally Posted by youngb11
I'm not sure if I understand how f'(\sqrt{x^2+9}) works.

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

Fernando Revilla