Answer
\[(a)F'(x)=\;\alpha x^{\alpha-1}f'(x^{\alpha})\]
\[(b)\;G'(x)=\alpha f'(x)[f(x)]^{\alpha-1}\: \]
Work Step by Step
\[(a) \; F(x)=f(x^{\alpha})\]
Differentiate with respect to $x$ using chain rule
\[ F'(x)=f'(x^{\alpha})\cdot \;(x^{\alpha})'\]
\[ F'(x)=f'(x^{\alpha})\cdot \;(\alpha x^{\alpha-1})\]
\[F'(x)=\alpha x^{\alpha-1}f'(x^{\alpha})\]
Hence,
\[F'(x)=\alpha x^{\alpha-1}f'(x^{\alpha})\]
\[(b)\; G(x)=[f(x)]^{\alpha}\]
Differentiate with respect to $x$ using chain rule
Because $\alpha$ is real number
\[G'(x)=\alpha[f(x)]^{\alpha-1}\: \{f(x)\}'\]
\[G'(x)=\alpha f'(x)[f(x)]^{\alpha-1}\: \]
Hence,
\[G'(x)=\alpha f'(x)[f(x)]^{\alpha-1}\: \]
