Miscellaneous · Q16: Implicit $\frac{dy}{dx}$ for $\cos y=x\cos(a+y)$⁠: $\frac{\cos^2(a+y)}{\sin a}$

Basics you need

Solution

1. What the condition $\cos a \ne \pm1$ guarantees

   (i) $\sin a \ne 0$⁠. Since $\sin^2 a = 1 - \cos^2 a$⁠, $\sin a = 0$ would force $\cos a = \pm 1$⁠, which is excluded.

   (ii) $\cos(a + y) \ne 0$ at every point of the curve. Suppose $\cos(a + y) = 0$ at some point. The equation then gives $\cos y = x\cdot 0 = 0$⁠, so $\sin y = \pm 1$⁠. Expanding,

   $$\begin{aligned} \cos(a + y) &= \cos a\cos y - \sin a\sin y \\ &= 0 - (\pm\sin a) = \mp\sin a \ne 0, \end{aligned}$$

   a contradiction. So on the curve we may divide by $\cos(a + y)$⁠:

   $$x = \frac{\cos y}{\cos(a + y)}.$$
   The answer has $\sin a$ in a denominator, and the working will divide by $\cos(a + y)$⁠. Part (i) shows the first is allowed; part (ii) shows the second is automatic once $\sin a \ne 0$⁠. This is why the question insists on $\cos a \ne \pm 1$⁠. If $\cos a = 1$ (say $a = 0$⁠), the equation becomes $\cos y = x\cos y$⁠, which is not a curve with a slope at all (it is the line $x = 1$ together with the horizontal lines where $\cos y = 0$⁠).
   $\cos(A + B)$ formula (Basic 3); $\cos y = 0 \Rightarrow \sin y = \pm 1$ (Basic 4). Basics: trigonometric identities

2. Differentiate both sides with respect to $x$

   Left side, chain rule with outer $\cos$ and inner $y$⁠: $\frac{d}{dx}\cos y = -\sin y\,\frac{dy}{dx}$⁠.

   Right side, product rule on $x\cdot\cos(a + y)$⁠, with the chain rule for $\cos(a+y)$ (inner $a + y$⁠, whose derivative is $\frac{dy}{dx}$ since $a$ is constant):

   $$\begin{aligned} \frac{d}{dx}\big[x\cos(a+y)\big] &= 1\cdot\cos(a+y) \\ &\quad - x\sin(a+y)\frac{dy}{dx}. \end{aligned}$$

   So

   $$-\sin y\,\frac{dy}{dx} = \cos(a+y) - x\sin(a+y)\,\frac{dy}{dx}.$$
   Every term containing $y$ picks up a factor $\frac{dy}{dx}$⁠, because $y$ depends on $x$⁠. The $x$ in front is differentiated normally (derivative 1). Implicit differentiation assumes $y$ is a differentiable function of $x$⁠; Step 5 shows why that is true here.
   Chain rule, Theorem 4; product rule, rule (2) of Section 5.3; implicit differentiation, Section 5.3.2. Lesson: implicit differentiation

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

   Move $x\sin(a+y)\frac{dy}{dx}$ to the left:

   $$\frac{dy}{dx}\big[x\sin(a+y) - \sin y\big] = \cos(a+y).$$
   This is the standard move in implicit differentiation: all terms with $\frac{dy}{dx}$ on one side, factor it out, everything else on the other side. Before dividing we need the bracket to be non-zero; the next step computes it exactly.
   Solving a linear equation for the unknown $\frac{dy}{dx}$⁠. Basics: solving for dy/dx

4. Simplify the bracket with the subtraction formula

   Substitute $x = \frac{\cos y}{\cos(a+y)}$ from Step 1:

   $$\begin{aligned} & x\sin(a+y) - \sin y \\ &= \frac{\cos y\sin(a+y) - \sin y\cos(a+y)}{\cos(a+y)} \\ &= \frac{\sin\big((a+y) - y\big)}{\cos(a+y)} = \frac{\sin a}{\cos(a+y)}. \end{aligned}$$

   This is non-zero (Step 1), so dividing Step 3 by it:

   $$\frac{dy}{dx} = \cos(a+y)\cdot\frac{\cos(a+y)}{\sin a} = \frac{\cos^2(a+y)}{\sin a}.$$
   The numerator $\sin A\cos B - \cos A\sin B$ with $A = a + y$⁠, $B = y$ is $\sin(A - B)$⁠, and $A - B = a$⁠: the variable $y$ cancels inside the sine. That is the whole trick of the question.
   $\sin(A - B) = \sin A\cos B - \cos A\sin B$ (Basic 3). Basics: trigonometric identities

5. Confirm through $\frac{dx}{dy}$

   From Step 1, $x$ is an explicit function of $y$⁠. By the quotient rule,

   $$\begin{aligned} \frac{dx}{dy} &= \frac{-\sin y\cos(a+y) + \cos y\sin(a+y)}{\cos^2(a+y)} \\ &= \frac{\sin a}{\cos^2(a+y)}. \end{aligned}$$

   This is never 0. Treating $y$ as the parameter ($x = x(y)$⁠, $y = y$⁠) the formula of Section 5.6 gives

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dy}{dx/dy} = \frac{1}{\sin a/\cos^2(a+y)} \\ &= \frac{\cos^2(a+y)}{\sin a}. \ ✓ \end{aligned}$$
   The same subtraction formula appears, and the answer matches. This route also shows that $y$ really is a differentiable function of $x$⁠: $\frac{dx}{dy}$ keeps one sign (that of $\sin a$⁠) on each interval where $\cos(a + y) \ne 0$⁠, so $x$ is strictly increasing or strictly decreasing in $y$ there (the sign test for increasing and decreasing functions, Chapter 6, used here without proof), and each $x$ corresponds to exactly one $y$ on that piece of the curve.
   Quotient rule (rule (3), Section 5.3); $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ with $\frac{dx}{dt} \ne 0$ (Section 5.6). Lesson: parametric differentiation

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