$\displaystyle \displaystyle y=f(\sqrt{x^2+9})$ 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!
$\displaystyle \displaystyle y=f(\sqrt{x^2+9})$ 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!
Use the Chain's Rule to $\displaystyle y=f(\sqrt{x^2+9})$ and substitute $\displaystyle x=4$ .
Fernando Revilla
$\displaystyle \dfrac{dy}{dx}=f'(\sqrt{x^2+9})\cdot \dfrac{x}{\sqrt{x^2+9}}$
Substitute $\displaystyle x=4$ .
Fernando Revilla
$\displaystyle \dfrac{dy}{dx}(4)=f'(\sqrt {4^2+9})\cdot \dfrac{4}{\sqrt {4^2+9}}=f'(5)\cdot \dfrac{4}{5}=(-2)\cdot \dfrac {4}{5}=-\dfrac{8}{5}$
Fernando Revilla