## Calculus: Early Transcendentals 8th Edition

$F'(x)=4(5x^{6}+2x^{3})^{3}(30x^{5}+6x^{2})$
$F(x)=(5x^{6}+2x^{3})^{4}$ Differentiate using the chain rule, which is $\dfrac{df(u)}{dx}=\dfrac{df}{du}\dfrac{du}{dx}$: $F'(x)=4(5x^{6}+2x^{3})^{3}(5x^{6}+2x^{3})'=...$ $...=4(5x^{6}+2x^{3})^{3}(30x^{5}+6x^{2})$