202212211306 Solution to 1987-CE-AMATH-I-2

Let $x=y+\sin y$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ and $\displaystyle{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}$ in terms of $y$ .

Roughwork.

$\begin{aligned} x & = y+\sin y \\ \mathrm{d}x & = \mathrm{d}y+\cos y\,\mathrm{d}y \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{1+\cos y} \\ \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = \frac{1}{(1+\cos y)^2} \\ \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 2\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\bigg(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg) \\ \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{1+\cos y}{2}\cdot \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{1}{(1+\cos y)^2}\bigg) \\ \end{aligned}$

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{1}{(1+\cos y)^2}\bigg) \\ &= (-2)(1+\cos y)^{-3}(-\sin y)\bigg( \frac{\mathrm{d}y}{\mathrm{d}x}\bigg) \\ &= (-2)(1+\cos y)^{-3}(-\sin y)\bigg(\frac{1}{1+\cos y}\bigg) \\ & = \frac{2\sin y}{(1+\cos y)^4} \\ \end{aligned}$

$\begin{aligned} \quad\enspace \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{1+\cos y}{2}\cdot \frac{2\sin y}{(1+\cos y)^4} \\ & = \frac{\sin y}{(1+\cos y)^3} \\ \end{aligned}$

Stretching the focus, compromise the better for it.

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