Exercise 5.5 · Q14: $\frac{dy}{dx}$ if $(\cos x)^y=(\cos y)^x$

Basics you need

Solution

1. Take logarithms of both sides

   Where $\cos x > 0$ and $\cos y > 0$⁠, both sides are positive and

   $$y\log\cos x = x\log\cos y.$$
   $\log(\cos x)^y = y\log\cos x$ and $\log(\cos y)^x = x\log\cos y$⁠. The book does not state the condition, but without $\cos x > 0$ and $\cos y > 0$ the powers with variable exponents (and their logs) are not defined.
   $\log_b p^n = n\log_b p$ (Section 5.4); positivity requirement (Section 5.5). Basics: exponentials and logarithms

2. Differentiate both sides with respect to $x$

   Left side, product rule on $y\cdot\log\cos x$⁠:

   $$\begin{aligned} &\frac{dy}{dx}\log\cos x + y\cdot\frac{-\sin x}{\cos x} \\ &= \frac{dy}{dx}\log\cos x - y\tan x. \end{aligned}$$

   Right side, product rule on $x\cdot\log\cos y$⁠:

   $$\begin{aligned} &\log\cos y + x\cdot\frac{-\sin y}{\cos y}\frac{dy}{dx} \\ &= \log\cos y - x\tan y\,\frac{dy}{dx}. \end{aligned}$$

   So

   $$\begin{aligned} &\frac{dy}{dx}\log\cos x - y\tan x \\ &= \log\cos y - x\tan y\,\frac{dy}{dx}. \end{aligned}$$
   $\log\cos x$ is a composite: outer $\log$⁠, inner $\cos x$⁠, so its derivative is $\frac{1}{\cos x}\cdot(-\sin x)$⁠. For $\log\cos y$ there are three layers: $\log$⁠, $\cos$⁠, and $y$ itself, so the chain rule gives $\frac{1}{\cos y}\cdot(-\sin y)\cdot\frac{dy}{dx}$⁠.
   Chain rule (Theorem 4, Section 5.3.1); $(\cos x)' = -\sin x$ (Table 5.3); $(\log x)' = \frac1x$ (Theorem 5). Lesson: implicit differentiation

3. Collect the $\frac{dy}{dx}$ terms

   $$\left(\log\cos x + x\tan y\right)\frac{dy}{dx} = \log\cos y + y\tan x.$$
   Move $-x\tan y\,\frac{dy}{dx}$ to the left (it becomes $+$⁠) and $-y\tan x$ to the right (it becomes $+$⁠).
   Rearranging a linear equation. Basics: solving for dy/dx

4. Solve for $\frac{dy}{dx}$⁠, with the condition

   $$\frac{dy}{dx} = \frac{y\tan x + \log\cos y}{x\tan y + \log\cos x},$$

   where $\cos x > 0$⁠, $\cos y > 0$ and $x\tan y + \log\cos x \ne 0$⁠.

   Dividing needs a non-zero divisor. Writing the terms in this order gives exactly the book's form.
   Division by a non-zero quantity. Basics: solving for dy/dx

5. Test the formula on the simplest pieces of the relation

   • The line $y = x$ (with $\cos x > 0$⁠): $(\cos x)^x = (\cos x)^x$ holds. The formula gives $\frac{x\tan x + \log\cos x}{x\tan x + \log\cos x} = 1$ wherever this is not $\frac00$⁠, the slope of the line.
   • The axis $y = 0$: $(\cos x)^0 = 1 = 1^x$ holds. The formula gives $\frac{0 + \log 1}{0 + \log\cos x} = 0$ wherever $\log\cos x \ne 0$⁠, the slope of the axis.
   • The axis $x = 0$: $1^y = 1 = (\cos y)^0$ holds. The denominator is $0 + \log 1 = 0$⁠, so the formula gives nothing, and rightly so: this line is vertical and has no finite slope.
   These quick cases show that the formula behaves sensibly, including failing exactly where a vertical piece of the relation appears. The relation also has other pieces, away from these lines; the check below tests the formula at a point of one of them.
   The slope of a line $y = mx + c$ is $m$⁠; a vertical line has no slope. Basics: slope at a point

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