Example 28: Derivative of $a^x$ is $a^x\log a$

Basics you need

Solution

1. $y$ is positive, so we may take its logarithm

   Let $y = a^x$⁠. Since $a > 0$⁠, $a^x > 0$ for every real $x$⁠. So $\log y$ exists, and by the power law for logs

   $$\log y = \log(a^x) = x\log a.$$
   Logarithmic differentiation needs the function to be positive (Section 5.5), and here it is, for every $x$⁠. Note that $\log a$ is a fixed number (for example $\log 2 \approx 0.6931$⁠): the right side is a constant multiple of $x$⁠, the simplest thing to differentiate.
   $\log p^n = n\log p$ with $p = a$⁠, $n = x$ (Section 5.4). Basics: exponentials and logarithms

2. Differentiate both sides with respect to $x$

   Left: chain rule with outer $\log$⁠, inner $y$⁠. Right: $\log a$ is a constant.

   $$\frac1y\cdot\frac{dy}{dx} = \log a.$$
   $\frac{d}{dx}(x\cdot c) = c$ for a constant $c$⁠; here $c = \log a$⁠. The common slip is to treat $\log a$ as if it depended on $x$⁠. It does not: $a$ is a constant.
   $\frac{d}{dx}\log y = \frac1y\frac{dy}{dx}$ (Theorem 5 and the chain rule). Lesson: logarithmic differentiation

3. Solve for $\frac{dy}{dx}$

   $$\frac{dy}{dx} = y\log a = a^x\log a.$$
   Multiply both sides by $y$ (allowed, as $y \ne 0$⁠) and substitute $y = a^x$ back.
   Rearranging an equation. Basics: solving for dy/dx

4. Method 2: write $a^x$ as a power of $e$

   Since $a = e^{\log a}$ (Example 25, valid as $a > 0$⁠), we have $a^x = \left(e^{\log a}\right)^x = e^{x\log a}$⁠. Now use the chain rule with inner $t = x\log a$ (so $\frac{dt}{dx} = \log a$⁠) and outer $e^t$⁠:

   $$\begin{aligned} \frac{d}{dx}\left(a^x\right) &= \frac{d}{dx}\left(e^{x\log a}\right) \\ &= e^{x\log a}\cdot\frac{d}{dx}(x\log a) \\ &= e^{x\log a}\cdot\log a \\ &= a^x\log a. \end{aligned}$$
   This is the book's "Alternatively". It shows that every exponential function is really $e^x$ with a stretched input: $a^x = e^{(\log a)x}$⁠. The stretch factor $\log a$ comes out as the inner derivative. Both methods agree, as they must.
   $(p^m)^n = p^{mn}$⁠; $\frac{d}{dx}e^t = e^t$ (Theorem 5); chain rule (Theorem 4). Lesson: exponential and logarithmic functions

5. What the factor $\log a$ tells us

   • $a = e$⁠: $\log e = 1$⁠, so $\frac{d}{dx}e^x = e^x$⁠, which is Theorem 5 again.
   • $a = 1$⁠: $\log 1 = 0$⁠, so the derivative is $0$⁠. Indeed $1^x = 1$ is constant.
   • $a > 1$⁠: $\log a > 0$⁠, so $a^x$ is increasing. $0 < a < 1$⁠: $\log a < 0$⁠, so $a^x$ is decreasing.
   At $x = 0$⁠, $a^0 = 1$⁠, so the slope there is exactly $\log a$⁠. Everywhere else the slope is $\log a$ times the height $a^x$⁠: an exponential function grows (or shrinks) at a rate proportional to its own size.
   $\log a > 0 \iff a > 1$⁠, because $\log$ is increasing with $\log 1 = 0$ (Section 5.4). Basics: exponentials and logarithms

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