Exercise 5.5 · Q13: $\frac{dy}{dx}$ if $y^x=x^y$

Basics you need

Solution

1. Take logarithms of both sides

   For $x > 0$⁠, $y > 0$ both sides are positive, so

   $$\begin{gathered} \log(y^x) = \log(x^y), \\ \text{that is}\quad x\log y = y\log x. \end{gathered}$$
   Unlike Q12, there is no sum here: each side is one power, so the log of each side can be simplified with $\log p^n = n\log p$⁠. The exponents $x$ and $y$ come down as ordinary factors.
   $\log_b p^n = n\log_b p$ for $p > 0$ (Section 5.4). Basics: exponentials and logarithms

2. Differentiate both sides with respect to $x$

   Each side is a product; use the product rule, and the chain rule for $\log y$⁠:

   $$\log y + x\cdot\frac1y\frac{dy}{dx} = \frac{dy}{dx}\log x + y\cdot\frac1x.$$
   Left: $x\log y$ gives $1\cdot\log y + x\cdot(\log y)'$⁠, and $(\log y)' = \frac1y\frac{dy}{dx}$ because $y$ depends on $x$⁠. Right: $y\log x$ gives $\frac{dy}{dx}\cdot\log x + y\cdot\frac1x$⁠.
   Product rule; chain rule; $(\log x)' = \frac1x$ (Theorem 5). Lesson: implicit differentiation

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

   $$\left(\frac{x}{y} - \log x\right)\frac{dy}{dx} = \frac{y}{x} - \log y.$$
   Move $\frac{dy}{dx}\log x$ to the left and $\log y$ to the right, changing their signs. The equation is linear in $\frac{dy}{dx}$⁠.
   Rearranging a linear equation. Basics: solving for dy/dx

4. Solve and clear the small fractions

   Where $\frac{x}{y} - \log x \ne 0$⁠,

   $$\frac{dy}{dx} = \frac{\frac{y}{x} - \log y}{\frac{x}{y} - \log x}.$$

   Multiply numerator and denominator by $xy$ (which is positive):

   $$\begin{aligned} \frac{dy}{dx} &= \frac{y^2 - xy\log y}{x^2 - xy\log x} \\ &= \frac{y}{x}\cdot\frac{y - x\log y}{x - y\log x}. \end{aligned}$$
   $xy\cdot\frac{y}{x} = y^2$ and $xy\cdot\frac{x}{y} = x^2$⁠. Taking out $y$ from the top and $x$ from the bottom gives the book's form. The condition $\frac{x}{y} - \log x \ne 0$ is the same as $x - y\log x \ne 0$⁠.
   Multiplying top and bottom by the same non-zero number leaves a fraction unchanged. Basics: solving for dy/dx

5. Where on the curve is the denominator zero?

   Suppose $x - y\log x = 0$ at a point of the curve, so $y\log x = x$⁠. The curve gives $x\log y = y\log x = x$⁠, so $\log y = 1$ and $y = e$⁠. Then $e\log x = x$⁠. Since $x > 0$⁠, this needs $\log x > 0$⁠, so $x > 1$⁠, and it says $\frac{x}{\log x} = e$⁠. The function $\frac{x}{\log x}$ has derivative $\frac{\log x - 1}{(\log x)^2}$⁠, negative on $(1, e)$ and positive for $x > e$⁠, so its smallest value on $x > 1$ is $\frac{e}{1} = e$⁠, taken only at $x = e$⁠. Hence the only such point is

   $$(x, y) = (e, e).$$

   So the formula holds at every point of the curve except $(e, e)$⁠, where it becomes $\frac00$⁠.

   This is sharper than "wherever the denominator is non-zero": we have found exactly where it fails. The point $(e, e)$ is where the two pieces of the curve cross (Step 6 and the figure), so no single slope can be expected there.
   Sign of the derivative shows where a function decreases and increases (Theorem 1, Section 6.3, in the later Chapter 6; applied on closed intervals inside $(1, e]$ and $[e, \infty)$⁠). Basics: slope at a point

6. The two pieces of the curve, and the formula on each

   The line $y = x$⁠. Every point $(x, x)$⁠, $x > 0$⁠, satisfies $x^x = x^x$⁠. There $y - x\log y = x - x\log x$ and $x - y\log x = x - x\log x$⁠, so for $x \ne e$ the formula gives $\frac{x}{x}\cdot 1 = 1$⁠, the slope of the line.

   The other piece. If $y \ne x$⁠, write $y = tx$ with $t > 0$⁠, $t \ne 1$⁠. Then $x\log(tx) = tx\log x$ gives $\log t = (t - 1)\log x$⁠, so

   $$x = t^{\frac{1}{t-1}}, \qquad y = t^{\frac{t}{t-1}}.$$

   For $t = 2$ this is the point $(2, 4)$⁠: indeed $4^2 = 2^4 = 16$⁠. The formula gives

   $$\begin{aligned} \frac{dy}{dx} &= \frac{4}{2}\cdot\frac{4 - 2\log 4}{2 - 4\log 2} \\ &= \frac{4(1 - \log 2)}{1 - 2\log 2} = -3.1773989. \end{aligned}$$
   Dividing $x\log(tx) = tx\log x$ by $x$ gives $\log t + \log x = t\log x$⁠. For $t = \frac12$ the same formulas give $(4, 2)$⁠, the mirror image of $(2, 4)$ in the line $y = x$⁠; there the formula gives $-0.3147228$⁠, the reciprocal of $-3.1773989$⁠, as reflection in $y = x$ requires. As $t \to 1$ the second piece approaches $(e, e)$⁠.
   $\log(pq) = \log p + \log q$ (Section 5.4). Lesson: logarithmic differentiation

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