Answer
$$f'(x) = bx^ax^{b-1}g'(x^b)+ax^{a-1}g(x^b)$$
Work Step by Step
Given $$f (x)= x^ag(x^b) $$
Then
\begin{align*}
f'(x)&= x^ag'(x^b)(bx^{b-1}) + ax^{a-1}g(x^b) \\
&= bx^ax^{b-1}g'(x^b)+ax^{a-1}g(x^b)
\end{align*}
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