Exercise 5.4 · Q3: Differentiate $e^{x^3}$

Basics you need

Solution

1. Where is the function defined?

   $x^3$ is defined for every real $x$⁠, and $e^t$ is defined for every real $t$⁠. So $y = e^{x^3}$ is defined for all $x \in \R$⁠, and its values are all positive.

   There is no division, no logarithm and no root here, so there is nothing to exclude. Both links are also differentiable everywhere, so we expect a derivative at every real $x$⁠.
   Domain of the exponential function is $\R$⁠, range the positive reals (Section 5.4). Basics: exponentials and logarithms

2. Name the inner and outer functions, and differentiate each

   Inner: $t = x^3$⁠. Outer: $y = e^t$⁠.

   $$\frac{dt}{dx} = 3x^2, \qquad \frac{dy}{dt} = e^t.$$
   To compute $e^{x^3}$ on a calculator you cube $x$ first and then press $e^x$⁠: the job done first is the inner function. The inner derivative is the power rule with $n = 3$⁠; the outer derivative is Theorem 5 (1) written in the letter $t$⁠.
   Power rule (Table 5.3) and Theorem 5 (1), Section 5.4. Lesson: exponential and logarithmic functions

3. Multiply (chain rule) and substitute back

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} = e^t\cdot 3x^2 \\ &= 3x^2 e^{x^3}, \qquad x \in \R. \end{aligned}$$
   The outer derivative $e^t$ is evaluated at the inner value $t = x^3$⁠, which gives back $e^{x^3}$⁠. The factor $3x^2$ is never negative, and $e^{x^3} > 0$⁠, so the derivative is $\ge 0$ everywhere. (That this makes the function increasing is a result of Chapter 6, Section 6.3; here it simply agrees with $x^3$⁠, and so $e^{x^3}$⁠, growing with $x$⁠.) It is $0$ only at $x = 0$⁠, where the graph is momentarily flat.
   Chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule

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