Miscellaneous · Q3: Differentiate $(5x)^{3\cos 2x}$ by taking logarithms

Basics you need

Solution

1. Fix the domain: $x > 0$

   Let $y = (5x)^{3\cos 2x}$⁠. Here the base is $u(x) = 5x$ and the exponent is $v(x) = 3\cos 2x$⁠. The exponent is not a whole number in general (for example at $x = 1$ it is about $-1.248$⁠), so the power only makes sense for a positive base:

   $$5x > 0 \iff x > 0.$$

   For $x > 0$ we also have $y > 0$⁠, because a positive number raised to any real power is positive. So $\log y$ is defined.

   The book stresses that in logarithmic differentiation "$f(x)$ and $u(x)$ must always be positive", since their logarithms are taken (Section 5.5). Stating the domain first is what makes every later step legal.
   A real power $p^n$ with a non-integer $n$ is defined for $p > 0$⁠, and it is then positive. Basics: exponentials and logarithms

2. Take logarithms of both sides

   $$\log y = \log (5x)^{3\cos 2x} = 3\cos 2x\cdot\log 5x.$$
   The rule $\log p^n = n\log p$ brings the exponent down in front. The awkward "variable power" becomes an ordinary product of two functions, $3\cos 2x$ and $\log 5x$⁠, each of which we can differentiate.
   $\log_b p^n = n\log_b p$ for any real $n$⁠, Section 5.4. Basics: exponentials and logarithms

3. Differentiate the left side: $\frac{d}{dx}\log y$

   Here $\log y$ is a composite: the outer function is $\log$⁠, the inner function is $y$ (a function of $x$⁠). By the chain rule,

   $$\frac{d}{dx}(\log y) = \frac{1}{y}\cdot\frac{dy}{dx}.$$
   $\frac{d}{dy}\log y = \frac1y$ (Theorem 5), and the chain rule multiplies by the derivative of the inner function, $\frac{dy}{dx}$⁠, which is exactly what we want to find.
   Chain rule, Theorem 4, Section 5.3.1. Lesson: chain rule

4. Differentiate the right side by the product rule

   The two factors and their derivatives:

   $$\begin{aligned} \frac{d}{dx}(3\cos 2x) &= 3(-\sin 2x)\cdot 2 \\ &= -6\sin 2x, \\ \frac{d}{dx}(\log 5x) &= \frac{1}{5x}\cdot 5 = \frac{1}{x}. \end{aligned}$$

   Each line is a chain rule: in the first the inner function is $2x$ (derivative 2), in the second the inner function is $5x$ (derivative 5). Now the product rule:

   $$\begin{aligned} &\frac{d}{dx}\left(3\cos 2x\cdot\log 5x\right) \\ &= -6\sin 2x\log 5x + 3\cos 2x\cdot\frac1x \\ &= \frac{3\cos 2x}{x} - 6\sin 2x\log 5x. \end{aligned}$$
   The product rule is "derivative of the first times the second, plus the first times derivative of the second". A quick way to see $\frac{d}{dx}\log 5x = \frac1x$⁠: $\log 5x = \log 5 + \log x$⁠, and $\log 5$ is a constant. The 5 cancels, which surprises many students.
   Product rule (2), Section 5.3; $\frac{d}{dx}\cos x = -\sin x$ (Table 5.3); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Basics: derivative rules

5. Equate and multiply by $y$

   Steps 3 and 4 differentiate the two sides of the same equation, so they are equal:

   $$\frac1y\cdot\frac{dy}{dx} = \frac{3\cos 2x}{x} - 6\sin 2x\log 5x.$$

   Multiply both sides by $y = (5x)^{3\cos 2x}$⁠:

   $$\begin{aligned} \frac{dy}{dx} = {}&(5x)^{3\cos 2x}\Bigl[\frac{3\cos 2x}{x} \\ &\qquad - 6\sin 2x\log 5x\Bigr]. \end{aligned}$$
   The derivative must be written in terms of $x$⁠, so we put back $y = (5x)^{3\cos 2x}$⁠. Multiplying is allowed because $y$ is a number (non-zero, in fact positive), and nothing is divided by zero since $x > 0$⁠.
   Logarithmic differentiation, Section 5.5 (compare Example 29, $x^{\sin x}$⁠). Lesson: logarithmic differentiation

6. Cross-check by writing the power as an exponential

   For $x > 0$⁠, $5x = e^{\log 5x}$⁠, so $y = e^{3\cos 2x\log 5x}$⁠. By the chain rule with outer function $e^{(\ )}$ and inner function $3\cos 2x\log 5x$⁠:

   $$\frac{dy}{dx} = e^{3\cos 2x\log 5x}\cdot\frac{d}{dx}\left(3\cos 2x\log 5x\right).$$

   The first factor is $y$ itself and the second is the derivative of Step 4, so we get the same answer.

   Logarithmic differentiation and "write it as $e^{(\ )}$" are the same idea seen from two sides. Getting the same result two ways guards against slips.
   $\frac{d}{dx}e^x = e^x$ (Theorem 5); $p = e^{\log p}$ for $p > 0$ (Example 25). Basics: exponentials and logarithms

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