Example 30: $\frac{dy}{dx}$ if $y^x+x^y+x^x=a^b$

Basics you need

Solution

1. Name the three terms and differentiate the relation

   Put

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

   Then $u + v + w = a^b$ for every $x$ in question. The right side is a constant, so its derivative is $0$⁠, and by the sum rule

   $$\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0. \qquad (1)$$
   We cannot take the log of the whole equation usefully: $\log(u + v + w)$ does not split into $\log u + \log v + \log w$⁠. So we keep the sum, differentiate it term by term, and handle each power on its own. $a^b$ is a fixed number ($a$⁠, $b$ are constants), not a function of $x$⁠.
   Sum rule $(u + v + w)' = u' + v' + w'$ (algebra of derivatives, Section 5.3) and derivative of a constant is $0$ (Class XI). Lesson: implicit differentiation

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

   Since $y > 0$⁠, $u > 0$ and $\log u = x\log y$⁠. Differentiate: the right side is a product of $x$ and $\log y$⁠, and $\log y$ needs the chain rule because $y$ depends on $x$⁠:

   $$\begin{aligned} \frac1u\cdot\frac{du}{dx} &= x\cdot\frac{d}{dx}(\log y) + \log y\cdot\frac{d}{dx}(x) \\ &= x\cdot\frac1y\cdot\frac{dy}{dx} + \log y\cdot 1. \end{aligned}$$

   Multiply by $u = y^x$⁠:

   $$\frac{du}{dx} = y^x\left[\frac{x}{y}\,\frac{dy}{dx} + \log y\right]. \qquad (2)$$
   In $y^x$ both the base ($y$⁠, which depends on $x$⁠) and the exponent ($x$⁠) vary, so neither the power rule nor the $a^x$ rule applies on its own. Taking logs turns the power into the product $x\log y$⁠. The factor $\frac{dy}{dx}$ appears because we differentiate $\log y$ with respect to $x$⁠, not $y$⁠.
   $\log p^n = n\log p$ (Section 5.4); product rule; chain rule. Lesson: logarithmic differentiation

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

   Since $x > 0$⁠, $v > 0$ and $\log v = y\log x$⁠. Product rule again:

   $$\begin{aligned} \frac1v\cdot\frac{dv}{dx} &= y\cdot\frac{d}{dx}(\log x) + \log x\cdot\frac{dy}{dx} \\ &= \frac{y}{x} + \log x\cdot\frac{dy}{dx}. \end{aligned}$$

   Multiply by $v = x^y$⁠:

   $$\frac{dv}{dx} = x^y\left[\frac{y}{x} + \log x\,\frac{dy}{dx}\right]. \qquad (3)$$
   This time the factor that needs $\frac{dy}{dx}$ is $y$ itself, whose derivative with respect to $x$ is $\frac{dy}{dx}$⁠; and $\log x$ differentiates to $\frac1x$ directly.
   $\frac{d}{dx}\log x = \frac1x$ (Theorem 5); product rule (Section 5.3). Basics: derivative rules

4. $\frac{dw}{dx}$ for $w = x^x$

   $\log w = x\log x$⁠, so

   $$\frac1w\cdot\frac{dw}{dx} = x\cdot\frac1x + \log x\cdot 1 = 1 + \log x,$$ $$\frac{dw}{dx} = x^x(1 + \log x). \qquad (4)$$
   Here $y$ does not appear, so no $\frac{dy}{dx}$ arises. This result, $\frac{d}{dx}x^x = x^x(1 + \log x)$⁠, is worth remembering: it is a standard example of logarithmic differentiation.
   Product rule on $x\log x$⁠. Lesson: logarithmic differentiation

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

   Putting (2), (3), (4) into (1):

   $$\begin{aligned} &y^x\left[\frac{x}{y}\frac{dy}{dx} + \log y\right] + x^y\left[\frac{y}{x} + \log x\frac{dy}{dx}\right] \\ &\quad + x^x(1 + \log x) = 0. \end{aligned}$$

   Simplify the two mixed products with the index law: $y^x\cdot\frac{x}{y} = x\,y^{x-1}$ and $x^y\cdot\frac{y}{x} = y\,x^{y-1}$⁠. Keep the $\frac{dy}{dx}$ terms on the left and move the rest to the right:

   $$\begin{aligned} &\left(x\,y^{x-1} + x^y\log x\right)\frac{dy}{dx} \\ &= -\left[y^x\log y + y\,x^{y-1} + x^x(1 + \log x)\right]. \end{aligned}$$
   After implicit differentiation the equation is linear in $\frac{dy}{dx}$⁠. The terms containing it are $x y^{x-1}\frac{dy}{dx}$ (from $u'$⁠) and $x^y\log x\frac{dy}{dx}$ (from $v'$⁠); everything else is moved across with its sign changed.
   $\frac{y^x}{y} = y^{x-1}$ and $\frac{x^y}{x} = x^{y-1}$ for $x, y > 0$⁠. Basics: solving for dy/dx

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

   $$\frac{dy}{dx} = -\frac{N}{D},$$

   where

   $$\begin{aligned} N &= y^x\log y + y\,x^{y-1} + x^x(1 + \log x), \\ D &= x\,y^{x-1} + x^y\log x, \end{aligned}$$

   valid wherever the denominator $D = x\,y^{x-1} + x^y\log x \ne 0$⁠.

   We may only divide by a non-zero quantity. For $x \ge 1$ the denominator is certainly positive ($x y^{x-1} > 0$ and $\log x \ge 0$⁠). For $0 < x < 1$⁠, $\log x < 0$ and the denominator can vanish at some points of a curve of this family (for example it is zero at some $y$ between $1$ and $3$ when $x = 0.1$⁠); there the formula does not apply.
   Division requires a non-zero divisor. Basics: solving for dy/dx

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