Answer: (i) Differentiating \(\tan y=x\) with respect to \(x\) gives

\(\sec^2 y\,\frac{\mathrm d y}{\mathrm d x}=1\).

Since \(\tan^2 y= x^2\), we have \(\sec^2 y=1+\tan^2 y=1+x^2\), so

\((1+x^2)\frac{\mathrm d y}{\mathrm d x}=1\), hence \(\frac{\mathrm d y}{\mathrm d x}=\frac{1}{1+x^2}\).

Differentiate again:

\(\frac{\mathrm d}{\mathrm d x}\left((1+x^2)\frac{\mathrm d y}{\mathrm d x}\right)=0\), so

\(2x\frac{\mathrm d y}{\mathrm d x}+(1+x^2)\frac{\mathrm d^2 y}{\mathrm d x^2}=0\).

Therefore

\(\frac{\mathrm d^2 y}{\mathrm d x^2}=-\frac{2x}{1+x^2}\frac{\mathrm d y}{\mathrm d x}=-2x\left(\frac{\mathrm d y}{\mathrm d x}\right)^2\).

(ii) At \(\left(1,\frac{\pi}{4}\right)\),

\(\frac{\mathrm d y}{\mathrm d x}=\cos^2\left(\frac{\pi}{4}\right)=\frac12\).

So

\(\frac{\mathrm d^2 y}{\mathrm d x^2}=-2(1)\left(\frac12\right)^2=-\frac12\).

(i) Start from \(\tan y=x\). Differentiate implicitly with respect to \(x\):

\(\sec^2 y\,\frac{\mathrm d y}{\mathrm d x}=1\).

Using \(\sec^2 y=1+\tan^2 y\) and \(\tan y=x\), this becomes

\((1+x^2)\frac{\mathrm d y}{\mathrm d x}=1\).

Differentiate both sides again with respect to \(x\):

\(\frac{\mathrm d}{\mathrm d x}\left((1+x^2)\frac{\mathrm d y}{\mathrm d x}\right)=0\).

Apply the product rule:

\(2x\frac{\mathrm d y}{\mathrm d x}+(1+x^2)\frac{\mathrm d^2 y}{\mathrm d x^2}=0\).

Rearranging gives

\(\frac{\mathrm d^2 y}{\mathrm d x^2}=-\frac{2x}{1+x^2}\frac{\mathrm d y}{\mathrm d x}\).

From \(\frac{\mathrm d y}{\mathrm d x}=\frac{1}{1+x^2}\), we get

\(\frac{\mathrm d^2 y}{\mathrm d x^2}=-2x\left(\frac{1}{1+x^2}\right)^2=-2x\left(\frac{\mathrm d y}{\mathrm d x}\right)^2\).

(ii) At \(\left(1,\frac{\pi}{4}\right)\),

\(\frac{\mathrm d y}{\mathrm d x}=\cos^2\left(\frac{\pi}{4}\right)=\frac12\).

Substitute into the formula from part (i):

\(\frac{\mathrm d^2 y}{\mathrm d x^2}=-2(1)\left(\frac12\right)^2=-\frac12\).