Solution. (a) Quotient rule. Take \(u=3x-2\) and \(v=x+4\), so \(u'=3\) and \(v'=1\): \[\frac {dy}{dx}=\frac {u'v-uv'}{v^2}=\frac {3(x+4)-(3x-2)(1)}{(x+4)^2} =\frac {3x+12-3x+2}{(x+4)^2}=\frac {14}{(x+4)^2}.\] The \(x\) terms cancel, leaving a derivative that is positive everywhere — the curve is always increasing, which is worth noticing as a check.

(b) Chain rule, twice. Write \(y=\left [\cos (3x-1)\right ]^4\). The outermost operation is raising to the fourth power, then the cosine, then the linear inside. \[\frac {dy}{dx}=4\left [\cos (3x-1)\right ]^3\cdot \left (-\sin (3x-1)\right )\cdot 3 =-12\cos ^3(3x-1)\sin (3x-1).\] Work from the outside in, and multiply by the derivative of the inside at each stage. The factor of \(3\) at the end comes from differentiating \(3x-1\), and forgetting it is the usual chain-rule error. The minus sign comes from differentiating the cosine.

(c) Product rule with a chain inside. Take \(u=x^3\) and \(v=e^{4x}\), so \(u'=3x^2\) and, by the chain rule, \(v'=4e^{4x}\): \[\frac {dy}{dx}=u'v+uv'=3x^2e^{4x}+x^3\left (4e^{4x}\right ) =3x^2e^{4x}+4x^3e^{4x}.\] Factorising makes it tidier and easier to use later: \[\frac {dy}{dx}=x^2e^{4x}\left (3+4x\right ).\]