Example 43: Differentiate $\sin^2 x$ with respect to $e^{\cos x}$

Basics you need

Solution

1. Translate the question

   Let $u = \sin^2 x$ and $v = e^{\cos x}$⁠. We want $\dfrac{du}{dv}$⁠. Since both are functions of $x$⁠, the chain rule gives $\dfrac{du}{dx} = \dfrac{du}{dv}\cdot\dfrac{dv}{dx}$⁠, so

   $$\frac{du}{dv} = \frac{du/dx}{dv/dx} \qquad\left(\frac{dv}{dx} \ne 0\right).$$
   We cannot differentiate "with respect to $e^{\cos x}$" directly with our rules, which are all with respect to a single variable. Going through $x$ turns the problem into two ordinary derivatives and one division. This is the same reasoning the textbook uses for parametric forms in Section 5.6.
   Section 5.6, with $x$ as the parameter. Lesson: parametric differentiation

2. Find $\frac{du}{dx}$

   $u = (\sin x)^2$⁠: outer function $t^2$⁠, inner function $t = \sin x$⁠.

   $$\frac{du}{dx} = 2\sin x\cdot\frac{d}{dx}(\sin x) = 2\sin x\cos x.$$
   $\sin^2 x$ means $(\sin x)^2$⁠, the square of the sine, so the outer function is squaring. The power rule gives $2t$ and the chain rule multiplies by $(\sin x)' = \cos x$⁠.
   Power rule and chain rule; $(\sin x)' = \cos x$ (Table 5.3). Basics: derivative rules

3. Find $\frac{dv}{dx}$

   $v = e^{\cos x}$⁠: outer function $e^w$⁠, inner function $w = \cos x$⁠.

   $$\frac{dv}{dx} = e^{\cos x}\cdot(-\sin x) = -\sin x\,e^{\cos x}.$$
   The exponential is its own derivative, evaluated at the same input $\cos x$⁠; the chain rule then multiplies by $(\cos x)' = -\sin x$⁠. This is Example 26 (iv) of the textbook.
   Theorem 5 (1), Section 5.4; chain rule. Example 26

4. Where is $\frac{dv}{dx} \ne 0$⁠?

   $e^{\cos x} > 0$ for every $x$⁠, so $\frac{dv}{dx} = -\sin x\,e^{\cos x}$ is zero exactly when $\sin x = 0$⁠, that is $x = n\pi$⁠, $n \in \Z$⁠. We work with $x \ne n\pi$⁠.

   We are about to divide by $\frac{dv}{dx}$ and then cancel $\sin x$⁠; both need $\sin x \ne 0$⁠. The textbook cancels without stating this, so we state it.
   An exponential is never zero; $\sin x = 0$ exactly at integer multiples of $\pi$⁠. Basics: trigonometric identities

5. Divide and simplify

   For $x \ne n\pi$⁠:

   $$\begin{aligned} \frac{du}{dv} &= \frac{2\sin x\cos x}{-\sin x\,e^{\cos x}} \\ &= -\frac{2\cos x}{e^{\cos x}} = -2\cos x\,e^{-\cos x}. \end{aligned}$$
   The common factor $\sin x$ cancels because it is not zero (Step 4). The minus sign comes from $\frac{dv}{dx}$⁠. Dividing by $e^{\cos x}$ is the same as multiplying by $e^{-\cos x}$⁠.
   $\frac{1}{e^w} = e^{-w}$ (laws of exponents). Basics: exponentials and logarithms

6. Check by writing $u$ in terms of $v$

   Since $v = e^{\cos x}$⁠, we have $\cos x = \log v$⁠, and $\sin^2 x = 1 - \cos^2 x$⁠. So

   $$u = 1 - (\log v)^2, \qquad \tfrac1e \le v \le e.$$

   Differentiating with respect to $v$ by the chain rule:

   $$\frac{du}{dv} = -2\log v\cdot\frac1v = -\frac{2\cos x}{e^{\cos x}},$$

   the same answer. This route also shows why $x = n\pi$ is excluded: there $\cos x = \pm1$⁠, so $v = e$ or $\frac1e$⁠, an end of the interval of possible $v$-values, where only a one-sided rate exists.

   Here the parameter can be removed completely, and the direct derivative agrees with Step 5. Agreement of two independent routes is the best check that no sign was lost.
   $\log(e^w) = w$⁠; $\sin^2 x + \cos^2 x = 1$⁠; $(\log v)' = \frac1v$ (Theorem 5). Basics: exponentials and logarithms

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