Q26: $x^y = e^{x - y}$⁠: prove $\frac{dy}{dx} = \frac{\log x}{(\log(xe))^2}$⁠, or $\frac{d^2y}{dx^2}$ for the cycloid

The unknown $y$ is in a power, so taking logs brings it down. The two log rules used are:

So $\log x^y = y\log x$ and $\log e^{x - y} = (x - y)\log e = x - y$⁠.

Then collect the $y$ terms on one side, take $y$ out as a common factor and divide. Now $y$ is an explicit function of $x$⁠.

$y$ is a quotient, so use the quotient rule:

Here $u = x$⁠, $u' = 1$⁠, $v = 1 + \log x$ and $v' = \frac1x$⁠.

On the top, $x\cdot\frac1x = 1$⁠, and $1 + \log x - 1 = \log x$⁠.

The answer we must reach has $\log(xe)$⁠, so write the 1 as $\log e$⁠. Then use the product rule for logs:

Put $x = e$⁠: the top is $\log e = 1$⁠, and inside the bracket $xe = e\cdot e = e^2$⁠.

By the power rule for logs, $\log e^2 = 2\log e = 2$⁠. So the bottom is $2^2 = 4$⁠.

Both $x$ and $y$ are given in terms of the parameter $\theta$⁠, so first differentiate each with respect to $\theta$ ($a$ is a constant):

For a curve given by a parameter, $\frac{dy}{dx}$ is the ratio of the two $\theta$-derivatives.

To simplify, use the half-angle formulas:

Cancel $2\sin\frac{\theta}{2}$⁠, and $\frac{\cos}{\sin}$ of the same angle is $\cot$⁠.

(First the $a$ on top and bottom cancels.)

$\cot\frac{\theta}{2}$ is a function of $\theta$⁠, but we differentiate with respect to $x$⁠. So use the chain rule:

$\frac{d}{d\theta}\cot\frac{\theta}{2} = -\operatorname{cosec}^2\frac{\theta}{2}\cdot\frac12$⁠, and $\frac{d\theta}{dx}$ is the reciprocal of $\frac{dx}{d\theta} = a(1 - \cos\theta)$⁠.

Watch out: $\frac{d^2y}{dx^2}$ is not $\frac{d^2y}{d\theta^2}$ divided by $\frac{d^2x}{d\theta^2}$⁠.

Again $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$⁠. Move the $a$ next to the $\frac12$ first:

Then gather everything:

Since $\frac{1}{\sin^2\frac{\theta}{2}} = \operatorname{cosec}^2\frac{\theta}{2}$⁠, the two $\operatorname{cosec}^2$ factors multiply to $\operatorname{cosec}^4\frac{\theta}{2}$⁠, and $2a \times 2 = 4a$⁠.