Exercise 5.5 · Q1: Differentiate $\cos x\cos 2x\cos 3x$

Basics you need

Solution

1. Where is $y$ positive, negative or zero?

   Let $y = \cos x\cos 2x\cos 3x$⁠. It is defined and differentiable for every real $x$⁠, being a product of differentiable functions. But it is not always positive. At $x = 0$⁠, $y = 1\cdot 1\cdot 1 = 1 > 0$⁠. At $x = 0.6$ (radians), $\cos 0.6 = 0.8253\ldots$⁠, $\cos 1.2 = 0.3624\ldots$ and $\cos 1.8 = -0.2272\ldots$⁠, so

   $$y = \cos 0.6\,\cos 1.2\,\cos 1.8 \approx -0.0679,$$

   which is negative.

   And $y = 0$ exactly when one of the cosines is zero:

   $$\begin{aligned} \cos x = 0 &\iff x = (2n+1)\tfrac{\pi}{2}, \\ \cos 2x = 0 &\iff x = (2n+1)\tfrac{\pi}{4}, \\ \cos 3x = 0 &\iff x = (2n+1)\tfrac{\pi}{6}, \end{aligned}$$

   with $n \in \mathbf{Z}$⁠. Call these the excluded points. Away from them, $y \ne 0$⁠, but $y$ may be negative, so $\log y$ may not exist.

   The book's method takes $\log y$⁠, which needs $y > 0$ ("as otherwise their logarithms are not defined", Section 5.5). Here $y$ changes sign, so we use $\log|y|$ instead: $|y| > 0$ at every point that is not excluded.
   $\cos\theta = 0$ exactly when $\theta$ is an odd multiple of $\frac{\pi}{2}$⁠. Basics: trigonometric identities

2. The tool: $\frac{d}{dx}\log|g(x)| = \frac{g'(x)}{g(x)}$ wherever $g(x) \ne 0$

   Let $g$ be differentiable with $g(x) \ne 0$⁠. Near such a point $g$ keeps one sign (it is continuous, Theorem 3).

   • If $g(x) > 0$⁠: $|g| = g$⁠, so $\frac{d}{dx}\log g = \frac{1}{g}\cdot g' = \frac{g'}{g}$⁠.
   • If $g(x) < 0$⁠: $|g| = -g > 0$⁠, so $\frac{d}{dx}\log(-g) = \frac{1}{-g}\cdot(-g')$ $= \frac{g'}{g}$⁠.

   Either way the result is $\frac{g'}{g}$⁠. In particular, $\frac{d}{dx}\log|y| = \frac{1}{y}\cdot\frac{dy}{dx}$⁠.

   This is what lets logarithmic differentiation work for functions that are negative in places: the absolute value makes the logarithm exist, and the derivative formula does not change. Each case is just the chain rule with outer function $\log$ and inner function $g$ or $-g$⁠.
   Chain rule (Theorem 4) with $\frac{d}{dt}\log t = \frac1t$ for $t > 0$ (Theorem 5). Lesson: logarithmic differentiation

3. Take $\log|y|$ and split it

   At any point that is not excluded, all three cosines are non-zero, so

   $$\begin{aligned} \log|y| &= \log\big(|\cos x|\,|\cos 2x|\,|\cos 3x|\big) \\ &= \log|\cos x| + \log|\cos 2x| \\ &\quad + \log|\cos 3x|. \end{aligned}$$
   The absolute value of a product is the product of the absolute values, and the three numbers $|\cos x|$⁠, $|\cos 2x|$⁠, $|\cos 3x|$ are positive, so the law $\log(pq) = \log p + \log q$ applies (twice). The product has become a sum.
   $\log_b pq = \log_b p + \log_b q$ (Section 5.4). Basics: exponentials and logarithms

4. Differentiate both sides

   Left side, by Step 2: $\frac1y\frac{dy}{dx}$⁠. Right side, by Step 2 on each term with $g = \cos kx$⁠, $g' = -k\sin kx$⁠:

   $$\begin{aligned} \frac{d}{dx}\log|\cos x| &= \frac{-\sin x}{\cos x} = -\tan x, \\ \frac{d}{dx}\log|\cos 2x| &= \frac{-2\sin 2x}{\cos 2x} = -2\tan 2x, \\ \frac{d}{dx}\log|\cos 3x| &= \frac{-3\sin 3x}{\cos 3x} = -3\tan 3x. \end{aligned}$$

   So

   $$\frac1y\cdot\frac{dy}{dx} = -\tan x - 2\tan 2x - 3\tan 3x.$$
   The factors $2$ and $3$ are the inner derivatives: $\cos 2x$ is a composite with outer $\cos$ and inner $2x$⁠, so its derivative is $-\sin 2x\cdot 2$⁠. Forgetting them is the most common error in this question.
   Chain rule: $\frac{d}{dx}\cos kx = -k\sin kx$⁠; $\frac{\sin\theta}{\cos\theta} = \tan\theta$⁠. Lesson: the chain rule

5. Multiply by $y$

   $$\begin{aligned} \frac{dy}{dx} &= -y\,(\tan x + 2\tan 2x + 3\tan 3x) \\ &= -\cos x\cos 2x\cos 3x \\ &\quad\times(\tan x + 2\tan 2x + 3\tan 3x). \end{aligned}$$

   This holds at every $x$ where $\cos x$⁠, $\cos 2x$ and $\cos 3x$ are all non-zero, that is, at every $x$ that is not an odd multiple of $\frac{\pi}{2}$⁠, $\frac{\pi}{4}$ or $\frac{\pi}{6}$⁠.

   Multiplying both sides by $y$ (non-zero here) isolates $\frac{dy}{dx}$⁠. This is exactly the textbook's answer. The restriction is real: at an excluded point one of the tangents is undefined, so this form cannot even be evaluated there.
   Rearranging: $\frac1y Y = R$ with $y \ne 0$ gives $Y = yR$⁠. Basics: solving for dy/dx

6. The product rule gives a form valid for every $x$

   With $u = \cos x$⁠, $v = \cos 2x$⁠, $w = \cos 3x$ and $u' = -\sin x$⁠, $v' = -2\sin 2x$⁠, $w' = -3\sin 3x$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= u'vw + uv'w + uvw' \\ &= -\sin x\cos 2x\cos 3x \\ &\quad - 2\cos x\sin 2x\cos 3x \\ &\quad - 3\cos x\cos 2x\sin 3x. \end{aligned}$$

   Where no cosine is zero, this equals the answer of Step 5: for instance $-\cos x\cos 2x\cos 3x\cdot\tan x$ $= -\sin x\cos 2x\cos 3x$⁠, since $\cos x\cdot\frac{\sin x}{\cos x} = \sin x$⁠; the other two terms match in the same way.

   The product rule needs no logarithms and no sign conditions, so this form is valid for every real $x$⁠. Expanding the bracket of Step 5 and cancelling one cosine in each term gives exactly these three terms, which confirms the textbook answer wherever it is defined.
   Product rule used twice: $(uvw)' = (uv)'w + (uv)w'$ $= u'vw + uv'w + uvw'$⁠. Basics: derivative rules

7. At an excluded point the derivative still exists

   Take $x = \frac{\pi}{6}$⁠, where $\cos 3x = \cos\frac{\pi}{2} = 0$⁠, so $y = 0$ and $\tan 3x$ is undefined. The product-rule form still works: the first two terms contain $\cos 3x = 0$⁠, so

   $$\begin{aligned} \frac{dy}{dx}\Big|_{x=\pi/6} &= -3\cos\tfrac{\pi}{6}\cos\tfrac{\pi}{3}\sin\tfrac{\pi}{2} \\ &= -3\cdot\tfrac{\sqrt3}{2}\cdot\tfrac12\cdot 1 = -\tfrac{3\sqrt3}{4}. \end{aligned}$$
   So the restriction in Step 5 belongs to the method (and the tan form of the answer), not to the function: $y$ is differentiable everywhere. $-\frac{3\sqrt3}{4} = -1.2990381\ldots$⁠, and a central difference quotient at $\frac{\pi}{6}$ gives the same.
   $\cos\frac{\pi}{6} = \frac{\sqrt3}{2}$⁠, $\cos\frac{\pi}{3} = \frac12$⁠, $\sin\frac{\pi}{2} = 1$⁠. Basics: trigonometric identities

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