Exercise 5.5 · Q12: $\frac{dy}{dx}$ if $x^y+y^x=1$

Basics you need

Solution

1. Name the two terms and differentiate the relation

   For $x > 0$⁠, $y > 0$ put

   $$u = x^y, \qquad v = y^x,$$

   so $u + v = 1$⁠. The right side is a constant, so by the sum rule

   $$\frac{du}{dx} + \frac{dv}{dx} = 0. \qquad (1)$$
   Both terms have a base and an exponent that vary with $x$ (remember $y$ depends on $x$⁠), so neither the power rule nor the $a^x$ rule applies; logs are needed. But $\log(u + v)$ does not split, so we differentiate the terms one at a time.
   Sum rule; derivative of a constant is 0 (Section 5.3). Lesson: implicit differentiation

2. $\frac{du}{dx}$ for $u = x^y$

   Since $x > 0$⁠, $u > 0$ and $\log u = y\log x$⁠. Product rule on $y\cdot\log x$⁠:

   $$\begin{aligned} \frac1u\frac{du}{dx} &= \frac{dy}{dx}\cdot\log x + y\cdot\frac1x. \end{aligned}$$

   Multiply by $u = x^y$⁠:

   $$\frac{du}{dx} = x^y\left[\frac{y}{x} + \log x\,\frac{dy}{dx}\right]. \qquad (2)$$
   The factor $y$ is a function of $x$⁠, so its derivative is $\frac{dy}{dx}$⁠; the factor $\log x$ differentiates to $\frac1x$ directly.
   $\log p^n = n\log p$ (Section 5.4); product rule; $(\log x)' = \frac1x$ (Theorem 5). Lesson: logarithmic differentiation

3. $\frac{dv}{dx}$ for $v = y^x$

   Since $y > 0$⁠, $v > 0$ and $\log v = x\log y$⁠. Product rule on $x\cdot\log y$⁠, with the chain rule for $\log y$⁠:

   $$\frac1v\frac{dv}{dx} = 1\cdot\log y + x\cdot\frac1y\frac{dy}{dx}.$$

   Multiply by $v = y^x$⁠:

   $$\frac{dv}{dx} = y^x\left[\log y + \frac{x}{y}\,\frac{dy}{dx}\right]. \qquad (3)$$
   This time the base is $y$⁠. Differentiating $\log y$ with respect to $x$ gives $\frac1y\cdot\frac{dy}{dx}$⁠: outer function $\log$⁠, inner function $y$⁠.
   Chain rule (Theorem 4, Section 5.3.1). Basics: derivative rules

4. Substitute into (1) and collect the $\frac{dy}{dx}$ terms

   Expanding (2) and (3) and using the index laws $x^y\cdot\frac{y}{x} = y\,x^{y-1}$ and $y^x\cdot\frac{x}{y} = x\,y^{x-1}$⁠, equation (1) becomes

   $$\begin{aligned} &y\,x^{y-1} + x^y\log x\,\frac{dy}{dx} \\ &+ y^x\log y + x\,y^{x-1}\frac{dy}{dx} = 0. \end{aligned}$$

   Keep the $\frac{dy}{dx}$ terms on the left:

   $$\begin{aligned} &\left(x^y\log x + x\,y^{x-1}\right)\frac{dy}{dx} \\ &= -\left(y\,x^{y-1} + y^x\log y\right). \end{aligned}$$
   After implicit differentiation the equation is linear in $\frac{dy}{dx}$⁠: two terms contain it, two do not. Moving the two that do not to the right changes their signs. $\frac{x^y}{x} = x^{y-1}$ because $x > 0$⁠.
   Index law $\frac{p^m}{p} = p^{m-1}$⁠. Basics: solving for dy/dx

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

   $$\frac{dy}{dx} = -\,\frac{y\,x^{y-1} + y^x\log y}{x^y\log x + x\,y^{x-1}},$$

   valid where $x^y\log x + x\,y^{x-1} \ne 0$⁠.

   We may only divide by a non-zero quantity. This is exactly the book's answer.
   Division needs a non-zero divisor. Basics: solving for dy/dx

6. A closer look: which points satisfy $x^y + y^x = 1$⁠?

   We show that $x^y + y^x > 1$ for all $x > 0$⁠, $y > 0$⁠.

   Case 1: $x \ge 1$ or $y \ge 1$⁠. If $x \ge 1$⁠, then $\log(x^y) = y\log x \ge 0$⁠, so $x^y \ge 1$⁠; and $y^x > 0$⁠. The sum is $> 1$⁠. The case $y \ge 1$ is the same with the letters swapped.

   Case 2: $0 < x < 1$ and $0 < y < 1$⁠. Fix $t$ with $0 < t < 1$ and let $\varphi(z) = 1 + tz - (1+z)^t$ for $z \ge 0$⁠. Then $\varphi(0) = 0$ and

   $$\begin{aligned} \varphi'(z) &= t\left[1 - (1+z)^{t-1}\right] \\ &> 0 \quad\text{for } z > 0, \end{aligned}$$

   because $(1+z)^{t-1} = e^{(t-1)\log(1+z)}$ and the exponent is negative ($t - 1 < 0$⁠, $\log(1+z) > 0$⁠), so $(1+z)^{t-1} < 1$⁠. Now take any $z > 0$⁠. On $[0, z]$⁠, $\varphi$ is continuous, and $\varphi' > 0$ inside, so by Theorem 1 of Chapter 6 (a later chapter, quoted here without proof) $\varphi$ is increasing on $[0, z]$⁠, and $\varphi(z) > \varphi(0) = 0$⁠:

   $$(1+z)^t < 1 + tz.$$

   Use this with $t = y$ and $1 + z = \frac1x$⁠, that is $z = \frac{1-x}{x} > 0$⁠:

   $$\begin{aligned} x^{-y} < 1 + \frac{y(1-x)}{x}, \text{ so} \\ x^y > \frac{x}{x + y(1-x)} > \frac{x}{x+y}. \end{aligned}$$

   (The last step uses $y(1-x) < y$⁠.) In the same way $y^x > \frac{y}{x+y}$⁠. Adding, $x^y + y^x > \frac{x+y}{x+y} = 1$⁠.

   So no point with $x > 0$⁠, $y > 0$ satisfies $x^y + y^x = 1$, and the calculation above is formal. It is still exactly the right calculation for every relation $x^y + y^x = c$ with a constant $c$⁠: the constant disappears in Step 1, so the same formula gives $\frac{dy}{dx}$⁠. For $c > 1$ these curves do exist; for $c = 2$ the point $(1, 1)$ is on the curve (see the check below).

   The book does not remark on this. The derivative of $(1+z)^t$ is $t(1+z)^{t-1}$ for any real $t$⁠, by logarithmic differentiation: $\log$ of it is $t\log(1+z)$⁠, whose derivative is $\frac{t}{1+z}$⁠. If a base were allowed to be $0$⁠, the half-lines $x = 0, y > 0$ and $y = 0, x > 0$ would satisfy the equation (since $0^y = 0$ and $y^0 = 1$⁠); but there no logarithm exists, and on $x = 0$ the relation does not even give $y$ as a function of $x$⁠.
   A function continuous on $[a, b]$ with positive derivative on $(a, b)$ is increasing on $[a, b]$ (Theorem 1, Section 6.3, in the later Chapter 6). Lesson: logarithmic differentiation

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