Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 18

Differentiating givesThe exponential never vanishes, so all stationary points have $x=k\pi /2$. At $x=k\pi$, $y=1$ and $y^{\prime \prime }=2>0$; at $x=\pi /2+k\pi$, $y=e$ and $y^{\prime \prime }=-2e<0$. ThusThese are global minima and maxima because $0\le {\sin }^{2}x\le 1$. None of the stationary points is an inflection point.

The graph is even and periodic with period $\pi$, bounded between $1$ and $e$. It rises from each minimum at $k\pi$ to a maximum halfway to the next minimum, then falls back. It has horizontal tangents at all these extrema and repeats smoothly.

For the composite square root, $y>1$ throughout the restricted $x$ interval. Differentiate with respect to $y$:The second expression comes from differentiating both $y^{-1}$ and $(\log y{)}^{-1/2}$. The endpoints where $y=1$ are excluded here.

On $0<x<\pi$, $\sqrt{\log y}=\sqrt{{\sin }^{2}x}=\sin x>0$. The chain rule givesUsing $\sin 2x=2\sin x\cos x$ yields

The second derivative chain rule isPut $s=\sin x>0$ and $c_0=\cos x$. Substituting the preceding results and $y^{\prime }=2ys{c}_0$ givesTherefore

Substitution first makes the square-root branch transparent:On the given interval this is exactly $\sin x$, so direct differentiation givesin agreement with the chain-rule calculation. Outside this interval one must retain the absolute value; there are cusps at integer multiples of $\pi$.