## Calculus: Early Transcendentals 8th Edition

(a) $$F'(x)=af'(x^a)x^{a-1}$$ (b) $$G'(x)=a[f(x)]^{a-1}f'(x)$$
(a) $$F(x)=f(x^a)$$ The derivative of $F(x)$ $$F'(x)=\frac{df(x^a)}{dx}=\frac{df(x^a)}{d(x^a)}\frac{d(x^a)}{dx}$$ (According to Chain Rule) We know that $\frac{d(x^a)}{dx}=ax^{a-1}$ So, $$F'(x)=af'(x^a)x^{a-1}$$ (b) $$G(x)=[f(x)]^a$$ The derivative of $G(x)$ $$G'(x)=\frac{d[f(x)]^a}{dx}=\frac{d[f(x)]^a}{df(x)}\frac{df(x)}{dx}$$ (according to Chain Rule) Now, $\frac{d[f(x)]^a}{df(x)}=a[f(x)]^{a-1}$ So, $$G'(x)=a[f(x)]^{a-1}f'(x)$$