Calculus (3rd Edition)

a) $2x\cos(x^{2}+1)$ b) $9x\sqrt {x^{2}+1}$ c) $4x^{3}+2x$
Using the chain rule, we have $\frac{d}{dx}(f(x^{2}+1))=\frac{d}{du}(f(u))\times\frac{d}{dx}(x^{2}+1)=f'(u)\times2x$ a) $f(u)=\sin u\implies f'(u)=\cos u$ $\frac{d}{dx}f(x^{2}+1)=\cos (u)\times2x$ $=2x\cos(x^{2}+1)$ b) $f(u)=3u^{3/2}\implies f'(u)=3\times\frac{3}{2}u^{1/2}=\frac{9}{2}\sqrt u$ $\frac{d}{dx}f(x^{2}+1)=\frac{9}{2}\sqrt u\times2x=9x\sqrt {x^{2}+1}$ c) $f(u)=u^{2}-u$, $f'(u)=2u-1$ $\frac{d}{dx}f(x^{2}+1)=(2u-1)2x$ $=[2(x^{2}+1)-1]2x=(2x^{2}+1)2x$ $=4x^{3}+2x$