Exercise 5.5 · Q15: $\frac{dy}{dx}$ if $xy=e^{x-y}$

Basics you need

Solution

1. Which points can lie on the curve?

   The right side $e^{x-y}$ is positive for all $x, y$⁠. So on the curve

   $$xy > 0:$$

   $x$ and $y$ are both non-zero and have the same sign. The curve has a piece with $x, y > 0$ (for example $(1, 1)$⁠, since $1 = e^0$⁠) and a piece with $x, y < 0$ (for example $(-1, -1)$⁠, since $1 = e^0$⁠).

   This tells us the logarithm of $xy$ always exists on the curve, even on the negative piece, where neither $\log x$ nor $\log y$ exists. It also tells us $x \ne 0$⁠, which we use when we divide later.
   $e^t > 0$ for every real $t$ (range of the exponential function, Section 5.4). Basics: exponentials and logarithms

2. Take logarithms of both sides

   Both sides are positive, so

   $$\log(xy) = \log e^{x-y} = x - y.$$
   The log undoes the exponential: $\log e^t = t$⁠. We keep $\log(xy)$ as the log of one positive number. On the positive piece one may write it as $\log x + \log y$ (that is what most solutions do), but on the negative piece that split is not allowed; keeping $\log(xy)$ works on both pieces.
   $\log e^t = t$ (Section 5.4). Lesson: logarithmic differentiation

3. Differentiate both sides with respect to $x$

   Left: with $g = xy$⁠, the product rule gives $g' = y + x\frac{dy}{dx}$⁠, so

   $$\frac{d}{dx}\log(xy) = \frac{y + x\frac{dy}{dx}}{xy}.$$

   Right: $\frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}$⁠. Hence

   $$\frac{y + x\frac{dy}{dx}}{xy} = 1 - \frac{dy}{dx}.$$
   $\log(xy)$ is a composite: outer $\log$⁠, inner $xy$⁠. Its derivative is $\frac{1}{xy}$ times the derivative of $xy$⁠, and $xy$ is a product of $x$ and $y$⁠, both depending on $x$⁠.
   Chain rule (Theorem 4, Section 5.3.1); product rule (Section 5.3). Lesson: implicit differentiation

4. Clear the fraction and collect the $\frac{dy}{dx}$ terms

   Multiply both sides by $xy$ (non-zero):

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

   Collect:

   $$\begin{aligned} (x + xy)\frac{dy}{dx} &= xy - y, \\ x(1 + y)\frac{dy}{dx} &= y(x - 1). \end{aligned}$$
   Moving $-xy\frac{dy}{dx}$ to the left and $y$ to the right; then take out the common factors $x$ on the left and $y$ on the right.
   Rearranging a linear equation; factorising. Basics: solving for dy/dx

5. Solve for $\frac{dy}{dx}$⁠, and find where it fails

   $$\frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)},$$

   valid where $x(y + 1) \ne 0$⁠. We know $x \ne 0$ on the curve, so the only problem is $y = -1$⁠. On the curve, $y = -1$ means $-x = e^{x+1}$⁠. The function $h(x) = -x - e^{x+1}$ has $h'(x) = -1 - e^{x+1} < 0$⁠, so it is decreasing and has at most one zero; $h(-1) = 1 - 1 = 0$⁠. So the only point of the curve with $y = -1$ is

   $$(-1, -1).$$

   There Step 4 reads $0\cdot\frac{dy}{dx} = -1\cdot(-2) = 2$⁠, which no number can satisfy: the curve has a vertical tangent at $(-1, -1)$ (see the figure).

   The formula holds at every point of the curve except $(-1, -1)$⁠. Turning it upside down, $\frac{dx}{dy} = \frac{x(y+1)}{y(x-1)}$ is $0$ at $(-1, -1)$⁠, which is what a vertical tangent means. At $(1, 1)$ the formula gives $\frac{1\cdot 0}{1\cdot 2} = 0$⁠: a horizontal tangent. On the positive piece $x$⁠, $y$ and $y + 1$ are positive, so $\frac{dy}{dx}$ has the sign of $x - 1$⁠: $y$ falls for $x < 1$ and rises for $x > 1$ (Theorem 1 of Chapter 6 again), which makes $(1, 1)$ the lowest point of that piece.
   A function with negative derivative is decreasing (Theorem 1, Section 6.3, in the later Chapter 6); division needs a non-zero divisor. Basics: slope at a point

6. A second route, without logs

   Differentiate $xy = e^{x-y}$ directly. Left: product rule, $y + x\frac{dy}{dx}$⁠. Right: chain rule with inner function $x - y$⁠:

   $$y + x\frac{dy}{dx} = e^{x-y}\left(1 - \frac{dy}{dx}\right).$$

   Replace $e^{x-y}$ by $xy$ (they are equal on the curve):

   $$y + x\frac{dy}{dx} = xy - xy\frac{dy}{dx},$$

   which is exactly the equation of Step 4. So both routes give the same $\frac{dy}{dx}$⁠.

   This confirms the answer, and shows what the logs did: they turned $e^{x-y}$ into $x - y$ before differentiating, instead of replacing it by $xy$ afterwards.
   $\frac{d}{dx}e^{u} = e^{u}\frac{du}{dx}$ (Theorem 5 with the chain rule). Basics: exponentials and logarithms

Every fact used, and where it comes from