Example 23: $\frac{dy}{dx}$ if $y+\sin y=\cos x$

Basics you need

Solution

1. The relation really gives one $y$ for each $x$⁠, and $|y| \le 1$

   Write $g(y) = y + \sin y$⁠, so the relation is $g(y) = \cos x$⁠. Two facts settle what $y$ can be.

   Size. Since $|\cos x| \le 1$⁠, any solution has $|y + \sin y| \le 1$⁠. If $|y| \le \pi$⁠, then $y$ and $\sin y$ have the same sign, so $|y + \sin y| = |y| + |\sin y|$⁠, which is $\ge |y|$⁠, giving $|y|\le 1$⁠. If $|y| > \pi$⁠, then $|y + \sin y| \ge |y| - 1$⁠, which is $> \pi - 1 > 1$⁠: impossible. So every solution satisfies

   $$-1 \le y \le 1.$$

   Exactly one solution. $g$ is continuous, takes the value $g(-1) \approx -1.84$ and $g(1) \approx 1.84$⁠, so by the intermediate value property it hits every number $\cos x$ in $[-1, 1]$ for some $y$ in $[-1, 1]$⁠. And $g$ is strictly increasing: for $y_1 < y_2$⁠,

   $$\begin{aligned} &g(y_2) - g(y_1) \\ &= (y_2 - y_1) + (\sin y_2 - \sin y_1) \\ &> 0, \end{aligned}$$

   because $|\sin y_2 - \sin y_1| < y_2 - y_1$ whenever $y_1 \ne y_2$⁠. So $g$ cannot take the same value twice, and each $x$ gives exactly one $y$⁠.

   The book takes it for granted that the relation defines $y$ as a function of $x$⁠. We check it because implicit differentiation only makes sense if there is a function $y(x)$ to differentiate. The bound $|y|\le1$ will also matter in Step 5. The inequality $|\sin y_2 - \sin y_1| < |y_2 - y_1|$ follows from the sum-to-product formula $$\sin y_2 - \sin y_1 = 2\cos\tfrac{y_1+y_2}{2}\,\sin\tfrac{y_2-y_1}{2},$$ since $|\cos| \le 1$ and $|\sin u| < |u|$ for $u \ne 0$⁠.
   Intermediate value property: a function continuous on $[a, b]$ takes every value between $f(a)$ and $f(b)$⁠. This chapter does not state or prove it, but it is a standard property of continuous functions; $g$ is continuous by the algebra of continuous functions (Theorem 1, Section 5.2.1). Basics: trigonometric identities

2. Differentiate every term with respect to $x$

   The relation holds for every $x$⁠, so both sides have equal derivatives:

   $$\frac{dy}{dx} + \frac{d}{dx}(\sin y) = \frac{d}{dx}(\cos x).$$

   Term by term:

   • $\frac{d}{dx}(y) = \frac{dy}{dx}$⁠: this is the unknown we want.
   • $\frac{d}{dx}(\sin y)$⁠: the outer function is $\sin$⁠, the inner function is $y(x)$⁠. By the chain rule, $\frac{d}{dx}(\sin y) = \cos y\cdot\frac{dy}{dx}$⁠.
   • $\frac{d}{dx}(\cos x) = -\sin x$⁠.

   So

   $$\frac{dy}{dx} + \cos y\cdot\frac{dy}{dx} = -\sin x.$$
   The most common error is to write $\frac{d}{dx}(\sin y) = \cos y$⁠. That would be the derivative with respect to $y$⁠. We are differentiating with respect to $x$⁠, and $y$ changes as $x$ changes, so the chain rule adds the factor $\frac{dy}{dx}$⁠: "derivative of the outer at the inner, times derivative of the inner".
   Chain rule (Theorem 4, Section 5.3.1) with $t = y(x)$ and $v(t) = \sin t$⁠: $\frac{d}{dx}\sin y = \cos y\cdot\frac{dy}{dx}$⁠. Lesson: chain rule

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

   $\frac{dy}{dx}$ appears in both terms on the left. Factor it out:

   $$\frac{dy}{dx}\,(1 + \cos y) = -\sin x.$$
   The equation is linear in the unknown $\frac{dy}{dx}$⁠: it appears only to the first power, multiplied by things we know ($1$ and $\cos y$⁠). That is always the case after implicit differentiation, so the next move is always to factor it out and divide.
   Factoring: $A + BA = A(1 + B)$⁠, here with $A = \frac{dy}{dx}$ and $B = \cos y$⁠. Basics: solving for dy/dx

4. Divide, and state when that is allowed

   We may divide by $1 + \cos y$ only if it is not zero:

   $$\frac{dy}{dx} = -\frac{\sin x}{1 + \cos y}, \qquad 1 + \cos y \ne 0.$$

   $1 + \cos y = 0$ means $\cos y = -1$⁠, which happens exactly when $y$ is an odd multiple of $\pi$⁠. So the formula holds where

   $$y \ne (2n+1)\pi, \qquad n \in \Z.$$
   This is the book's condition. Division by zero is undefined, so any formula obtained by dividing must carry the condition that the divisor is non-zero. Where $1+\cos y = 0$ the equation in Step 3 would read $0 = -\sin x$ and would say nothing about $\frac{dy}{dx}$⁠.
   $\cos\theta = -1 \iff \theta = (2n+1)\pi$ for an integer $n$ (unit circle: the point $(-1, 0)$⁠). Basics: trigonometric identities

5. On this curve the condition is never violated

   By Step 1, every point of the curve has $|y| \le 1$⁠. The odd multiples of $\pi$ closest to $0$ are $\pm\pi \approx \pm 3.14$⁠, far outside $[-1, 1]$⁠. In fact, for $|y|\le 1$⁠,

   $$\cos y \ge \cos 1 \approx 0.5403 > 0,$$

   so $1 + \cos y > 1.54$⁠. The formula $\frac{dy}{dx} = -\frac{\sin x}{1+\cos y}$ is therefore valid at every point of the curve, that is, for every real $x$⁠.

   The book states the condition $y \ne (2n+1)\pi$ but does not ask whether the curve ever reaches such a $y$⁠. Checking shows it does not: the restriction is correct but, for this particular relation, never bites. The slope is $0$ exactly where $\sin x = 0$⁠, at $x = n\pi$⁠: these are the highest and lowest points of the wave.
   $\cos$ decreases on $[0, \pi]$ and is even, so on $[-1, 1]$ its smallest value is $\cos 1$⁠. Basics: trigonometric identities

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