Answer: The curve has \(x=2\theta-\sin 2\theta\) and \(y=1-\cos 2\theta\).

So \(\frac{\mathrm{d}x}{\mathrm{d}\theta}=2-2\cos 2\theta=4\sin^2\theta\) and \(\frac{\mathrm{d}y}{\mathrm{d}\theta}=2\sin 2\theta=4\sin\theta\cos\theta\).

Hence, for \(\sin\theta\neq 0\),

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}=\frac{4\sin\theta\cos\theta}{4\sin^2\theta}=\cot\theta\).

This fails when \(\sin\theta=0\), that is, when \(\theta=k\pi\) for integer \(k\).

Also,

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\cot\theta\right)=\frac{\frac{\mathrm{d}}{\mathrm{d}\theta}(\cot\theta)}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}=\frac{-\csc^2\theta}{4\sin^2\theta}.\)

When \(\theta=\frac{\pi}{4}\), \(\sin\theta=\frac{\sqrt2}{2}\), so \(\sin^2\theta=\frac12\) and \(\csc^2\theta=2\). Therefore

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{-2}{4\cdot \frac12}=-1\).

Differentiate the parametric equations with respect to \(\theta\):

\(\frac{\mathrm{d}x}{\mathrm{d}\theta}=2-2\cos 2\theta=4\sin^2\theta\), using \(1-\cos 2\theta=2\sin^2\theta\).

\(\frac{\mathrm{d}y}{\mathrm{d}\theta}=2\sin 2\theta=4\sin\theta\cos\theta\).

Then

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}=\frac{4\sin\theta\cos\theta}{4\sin^2\theta}=\cot\theta\).

This is valid provided \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\neq 0\), so we need \(4\sin^2\theta\neq 0\). Hence the exceptional values are \(\theta=k\pi\), where \(k\) is an integer.

For the second derivative, use

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

Since \(\frac{\mathrm{d}y}{\mathrm{d}x}=\cot\theta\),

\(\frac{\mathrm{d}}{\mathrm{d}\theta}(\cot\theta)=-\csc^2\theta\).

So

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{-\csc^2\theta}{4\sin^2\theta}.\)

At \(\theta=\frac{\pi}{4}\), \(\sin\theta=\frac{\sqrt2}{2}\), so \(\csc^2\theta=2\) and \(4\sin^2\theta=2\). Therefore

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