Solution. \(y=\tan x = \frac {\sin x}{\cos x}\)

Let \(u=\sin x\) and \(v=\cos x\)

\(\implies \qquad \frac {du}{dx}=\cos x\,,\quad \frac {dv}{dx}=-\sin x\)

\begin {align*} \text {Then}\qquad \frac {dy}{dx} &=\frac {v\frac {du}{dx}-u\frac {dv}{dx}}{v^2}\\\\ & =\frac {\cos x (\cos x)-\sin x(-\sin x)}{\cos ^2x}\\\\ & = \frac {\cos ^2 x+\sin ^2 x}{\cos ^2x}\\\\ & = \frac {1}{\cos ^2x}\\\\ & = \sec ^2 x\\ \end {align*}

Solution. (i). With \(u=x^{4}-2x\) and \(v=x^{2}+1\), \[f'(x)=\frac {\left (x^{2}+1\right )\left (4x^{3}-2\right )-\left (x^{4}-2x\right )(2x)} {\left (x^{2}+1\right )^{2}} .\] Expanding the numerator, \[4x^{5}+4x^{3}-2x^{2}-2-2x^{5}+4x^{2}=2x^{5}+4x^{3}+2x^{2}-2,\] so \[f'(x)=\frac {2\left (x^{5}+2x^{3}+x^{2}-1\right )}{\left (x^{2}+1\right )^{2}} .\]

(ii). Simplify before differentiating — always worth a moment’s look. Since \(2x^{4}-x=x\left (2x^{3}-1\right )\), \[f(x)=\frac {3x^{2}}{x\left (2x^{3}-1\right )}=\frac {3x}{2x^{3}-1} \quad (x\neq 0),\] and now the quotient rule on a much smaller expression: \[f'(x)=\frac {\left (2x^{3}-1\right )(3)-3x\left (6x^{2}\right )}{\left (2x^{3}-1\right )^{2}} =\frac {-12x^{3}-3}{\left (2x^{3}-1\right )^{2}} .\] At \(x=\frac {1}{2}\) we have \(2x^{3}-1=\frac {1}{4}-1=-\frac {3}{4}\) and \(-12x^{3}-3=-\frac {3}{2}-3=-\frac {9}{2}\), so \[f'\left (\tfrac {1}{2}\right )=\frac {-\frac {9}{2}}{\left (-\frac {3}{4}\right )^{2}} =\frac {-\frac {9}{2}}{\frac {9}{16}}=-8 .\]