Exercise 5.3 · Q2: $\frac{dy}{dx}$ if $2x+3y=\sin y$

Basics you need

Solution

1. Is $y$ really a function of $x$⁠?

   We cannot solve for $y$⁠, but we can solve for $x$⁠: $x = \dfrac{\sin y - 3y}{2}$⁠. As $y$ increases, this $x$ strictly decreases, because $\frac{dx}{dy} = \frac{\cos y - 3}{2} < 0$ for every $y$⁠. It is continuous, and it tends to $-\infty$ as $y \to +\infty$ and to $+\infty$ as $y \to -\infty$⁠, so it takes every real value. So each real $x$ comes from exactly one $y$⁠: the relation defines $y$ as a function of $x$ on all of $\R$⁠.

   Section 5.3.2 takes it for granted that a relation like this defines $y$ as a differentiable function of $x$⁠. Here we can at least see that $y$ is a genuine function of $x$⁠: a strictly decreasing, continuous $x(y)$ taking every real value can be "read backwards". That $y(x)$ is also differentiable is assumed, as in the book. This step is reassurance, not a required part of the textbook method, and it uses two facts this chapter does not prove (see the note below).
   Two facts from outside this chapter: a negative derivative on an interval makes a function decreasing (Theorem 1, Section 6.3, Chapter 6), and a continuous function takes every value between two of its values (the intermediate value property, a standard property of continuous functions not proved in the book). A strictly decreasing function is one-one; one-one and onto $\R$⁠, it is invertible (Chapter 1, Section 1.4).

2. Differentiate both sides with respect to $x$

   $$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y).$$

   Term by term: $\frac{d}{dx}(2x) = 2$ and $\frac{d}{dx}(3y) = 3\frac{dy}{dx}$⁠. For the right side, the outer function is $\sin$ and the inner function is $y$⁠:

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

   So

   $$2 + 3\frac{dy}{dx} = \cos y\,\frac{dy}{dx}.$$
   $\sin y$ is not $\sin x$⁠: it is the sine of a quantity that itself changes with $x$⁠. The chain rule says: differentiate the sine with respect to its own input (giving $\cos y$⁠), then multiply by how fast that input changes with $x$ (giving $\frac{dy}{dx}$⁠).
   Chain rule, Theorem 4, with inner function $y(x)$ and outer function $\sin$⁠; used the same way in Example 23. Lesson: the chain rule

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

   $$\begin{aligned} 2 &= \cos y\,\frac{dy}{dx} - 3\frac{dy}{dx} \\ &= (\cos y - 3)\,\frac{dy}{dx}, \end{aligned}$$

   and therefore

   $$\frac{dy}{dx} = \frac{2}{\cos y - 3}.$$
   Subtract $3\frac{dy}{dx}$ from both sides so that all $\frac{dy}{dx}$ terms are together, factor out $\frac{dy}{dx}$⁠, and divide by the bracket. Division needs the bracket to be non-zero; the next step shows it always is.
   Factor out the unknown, then divide. Basics: solving for dy/dx

4. The divisor is never zero

   Since $\cos y \le 1$⁠, we have $\cos y - 3 \le -2 < 0$⁠. So $\frac{dy}{dx} = \frac{2}{\cos y - 3}$ is valid at every point of the curve, and in fact $-1 \le \frac{dy}{dx} \le -\frac12$⁠.

   The book's Example 23 ends with a condition on $y$ because its divisor $1 + \cos y$ can vanish. Here no such condition is needed. The bounds come from $-4 \le \cos y - 3 \le -2$⁠: dividing $2$ by a number between $-4$ and $-2$ gives a number between $-1$ and $-\frac12$⁠.
   $-1 \le \cos y \le 1$⁠. Basics: trigonometric identities

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