## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{5}{2}z^{3/2}+\dfrac{1}{2}z^{-1/2}e^{z}+z^{1/2}e^{z}$
$y=(z^{2}+e^{z})\sqrt{z}$ Rewrite the function like this: $y=(z^{2}+e^{z})z^{1/2}$ Differentiate using the product rule: $y'=(z^{2}+e^{z})(z^{1/2})'+(z^{1/2})(z^{2}+e^{z})'=...$ $...=(z^{2}+e^{z})(\dfrac{1}{2}z^{-1/2})+(z^{1/2})(2z+e^{z})=...$ Evaluate the products and simplify $...=\dfrac{1}{2}z^{3/2}+\dfrac{1}{2}z^{-1/2}e^{z}+2z^{3/2}+z^{1/2}e^{z}=...$ $...=\dfrac{5}{2}z^{3/2}+\dfrac{1}{2}z^{-1/2}e^{z}+z^{1/2}e^{z}$