Solution.

Let \(u=x^2\) and \(v=\cos x\), then

\(\frac {du}{dx}=2x,\qquad \frac {dv}{dx}=-\sin x\)

\(y=uv\)

\begin {align*} \frac {dy}{dx} & = v \frac {du}{dx}+u \frac {dv}{dx}\\\\ & = \cos x (2x) + (x^2) (-\sin x)\\\\ & = 2x\cos x -x^2\sin x\\ \end {align*}

Solution. (i). A product whose factors each need the chain rule. With \(u=\left (x^{3}-1\right )^{7}\) and \(v=\left (3x^{2}+2\right )^{5}\), \[\frac {du}{dx}=7\left (x^{3}-1\right )^{6}\left (3x^{2}\right )=21x^{2}\left (x^{3}-1\right )^{6}, \quad \frac {dv}{dx}=5\left (3x^{2}+2\right )^{4}(6x)=30x\left (3x^{2}+2\right )^{4},\] so \[\frac {dy}{dx}=21x^{2}\left (x^{3}-1\right )^{6}\left (3x^{2}+2\right )^{5} +30x\left (3x^{2}+2\right )^{4}\left (x^{3}-1\right )^{7} .\] Taking out the common factor \(3x\left (x^{3}-1\right )^{6}\left (3x^{2}+2\right )^{4}\), \[\frac {dy}{dx}=3x\left (x^{3}-1\right )^{6}\left (3x^{2}+2\right )^{4} \left [7x\left (3x^{2}+2\right )+10\left (x^{3}-1\right )\right ].\]

(ii). With \(u=x^{3}\), \(v=\tan x\): \[\frac {dy}{dx}=3x^{2}\tan x+x^{3}\sec ^{2}x .\]

(iii). With \(u=e^{-2x}\), \(v=\sin 3x\), each needing the chain rule: \[\frac {dy}{dx}=-2e^{-2x}\sin 3x+3e^{-2x}\cos 3x =e^{-2x}\left (3\cos 3x-2\sin 3x\right ).\]