## Calculus, 10th Edition (Anton)

$f'(x)=$ $37(x^3 + 2x)^{36}(3x^2+2)$
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. $f (x)=(g (x))^{37}$ $f'(x)=37 (g (x))^{37-1}×(g'(x))$ $f'(x)=37 (g (x))^{36}×(3x^{3-1}+2)$ $f'(x)=37 (x^3+2x)^{36}×(3x^2+2)$ $f'(x)=37 (x^3+2x)^{36}(3x^2+2)$