## Calculus 10th Edition

$y'=\dfrac{9x^2+4}{5\sqrt[5]{(3x^3+4x)^4}}.$ The derivative evaluated at the point $(2, 2)$ is $\dfrac{1}{2}.$
$u=3x^3+4x$; $\dfrac{du}{dx}=9x^2+4$ $y=u^{\frac{1}{5}};\dfrac{dy}{du}=\dfrac{1}{5\sqrt[5]{u^4}}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=\dfrac{9x^2+4}{5\sqrt[5]{(3x^3+4x)^4}}.$ To evaluate the derivative, plug in $x=2\rightarrow\dfrac{9(2)^2+4}{5\sqrt[5]{(3(2)^3+4(2))^4}}=\dfrac{1}{2}$ A graphing utility was used to verify this result.