Answer
$y'=\dfrac{x+4}{\sqrt{x^2+8x}}.$
The derivative at the point $(1, 3)$ is $\dfrac{5}{3}.$
Work Step by Step
Using the Chain Rule:
$u=x^2+8x$; $\dfrac{du}{dx}=2(x+4)$
$y=\sqrt{u};\dfrac{dy}{du}=\dfrac{1}{2\sqrt{u}}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=\dfrac{x+4}{\sqrt{x^2+8x}}.$
To evaluate the derivative, plug in $x=1\rightarrow\dfrac{1+4}{\sqrt{1+8}}=\dfrac{5}{3}.$
A graphing utility was used to verify this result.
