Answer
$f'(x)=$ $37(x^3 + 2x)^{36}(3x^2+2) $
Work Step by Step
Given that $f(x)=(x^3+2x)^{37} $
We can write this as $ f (x)=(g (x))^{37} $, where $ f (x)$ is the outside function and $ g (x)=x^3+2x $, the inside function.
Thus the answer is the derivative of the outside function, using the chain rule, times the derivative of the inside function.
$ f (x)=(g (x))^{37}$
$f'(x)=37 (g (x))^{37-1}×(g'(x)) $
$f'(x)=37 (g (x))^{36}×(3x^{3-1}+2) $
$f'(x)=37 (x^3+2x)^{36}×(3x^2+2) $
$f'(x)=37 (x^3+2x)^{36}(3x^2+2) $
