The variable in the exponent and logarithmic differentiation

Summary

This lesson differentiates the functions that have the variable in the exponent. All of them reduce to the natural exponential: $a^x = e^{x \ln a}$ , $x^r = e^{r \ln x}$ and $\left(f(x)\right)^{g(x)} = e^{g(x) \ln f(x)}$ , so that the derivative of $e^x$ , that of the logarithm and the chain rule suffice. One thus obtains the derivatives of the exponential and of the logarithm to any base, of powers with real exponents and of powers whose base and exponent both vary, such as $x^x$ or $x^{\sin x}$ . The derivative of $\ln \abs{f(x)}$ , which is $\frac{f'(x)}{f(x)}$ , turns products, quotients and powers into sums: this is logarithmic differentiation, which the lesson justifies without assuming what it wants to prove. At the end, the tangent to the logarithm through the origin decides which is larger, $\pi^e$ or $e^\pi$ , and explains why the outcome depends on the number $e$ .

Learning objectives

Differentiate the exponential and the logarithm to any base, and their compositions, starting from the equality $a^x = e^{x \ln a}$ .
Differentiate powers with real exponents and powers with variable base and exponent, writing them as exponentials with base $e$ , on the domain where the base is positive.
Apply logarithmic differentiation to products, quotients and powers, first checking that the function is differentiable and does not vanish at the point.
Compare powers such as $\pi^e$ and $e^\pi$ with the tangent to the logarithm and the fundamental inequality of the exponential.

The exponential and the logarithm to any base

The lesson on the derivatives of the elementary functions proved that $e^x$ is its own derivative and that the derivative of $\ln x$ is $\frac{1}{x}$ , and announced that the derivative of $a^x$ is $a^x \ln a$ . There is no need to repeat the computation with the difference quotient: every positive base is a power of $e$ . By the definition of the natural logarithm, from the lesson on the logarithm function, $e^{\ln y} = y$ for every $y > 0$ ; and by the law of the logarithm of a power, from the same lesson, $\ln\left(a^x\right) = x \ln a$ . With the chain rule, the rest is immediate.

TheoremDerivatives of the exponential and of the logarithm to base a

Let $a > 0$ , with $a \neq 1$ . The exponential $a^x$ is differentiable on $\mathbb{R}$ , the logarithm $\log_a x$ is differentiable on $\ioo{0}{\infty}$ , and the function $\log_a \abs{x}$ is differentiable on $\mathbb{R} \setminus \{0\}$ :

\frac{\mathrm{d}}{\mathrm{d}x}\left(a^x\right) = a^x \ln a, \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(\log_a x\right) = \frac{1}{x \ln a} \quad (x > 0), \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(\log_a \abs{x}\right) = \frac{1}{x \ln a} \quad (x \neq 0)

The exponential with base $a$ is written with the natural exponential and the chain rule is applied; the logarithm to base $a$ is a constant multiple of the natural logarithm. The figure shows $2^x$ , $e^x$ and $4^x$ with their tangents at the point $(0, 1)$ , with slopes $\ln 2$ , $1$ and $\ln 4$ , and at the end the tangent to $\log_2 x$ at $(1, 0)$ , the reflection of that of $2^x$ .

Proof

$a^x = e^{\ln\left(a^x\right)} = \dato{base}{e^{x \ln a}}$
Let $x$ be a real number. The number $a^x$ is positive; by the definition of the natural logarithm and the law of the logarithm of a power, from the lesson on the logarithm function:
$\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{x \ln a}\right) = e^{x \ln a} \cdot \ln a = \resaltar{a^x \ln a}$
The inner function $x \ln a$ is linear, with derivative $\ln a$ , and the outer function $e^t$ is differentiable on all of $\mathbb{R}$ , by the theorem on the derivative of the natural exponential. By the chain rule, from the lesson on the chain rule:
$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log_a \abs{x}\right) = \frac{1}{\ln a} \cdot \frac{1}{x} = \resaltar{\frac{1}{x \ln a}}$
By the change of base theorem, from the lesson on the logarithm function, $\log_a y = \frac{\ln y}{\ln a}$ for every $y > 0$ , with $\ln a \neq 0$ because $a \neq 1$ . By the theorem on the derivative of the natural logarithm, which holds for $\ln x$ and for $\ln \abs{x}$ , and the one on the derivative of a linear combination, from the lesson on differentiation rules:

The derivative of $a^x$ is proportional to the function, with the constant $\ln a$ , which is positive if $a > 1$ and negative if $0 < a < 1$ ; by the injectivity of the logarithm, it equals $1$ only if $a = e$ . With $a = 1$ , the function $1^x$ is constant and the formula still holds, because $\ln 1 = 0$ . At $x = 0$ , the definition of the derivative gives the limit $\frac{a^h - 1}{h} \to \ln a$ , which is the corollary on the standard limits of the logarithm and of powers, from the lesson on the continuity of the elementary functions: the derivative of $a^x$ at $0$ was already known, and the law $a^{x + h} = a^x a^h$ carries it over to every point.

CorollaryCompositions with the exponential and the logarithm

Let $u$ be a function defined on $A$ , differentiable at a point $a$ of $A$ that is an accumulation point of $A$ , and let $b > 0$ , with $b \neq 1$ . Then $e^{u(x)}$ and $b^{u(x)}$ are differentiable at $a$ . If $u(a) \neq 0$ , the functions $\ln \abs{u(x)}$ and $\log_b \abs{u(x)}$ are differentiable at $a$ , on their domain; and if $u(a) > 0$ , so is $\ln u(x)$ , with the same derivative as $\ln \abs{u(x)}$ . The derivatives at $x = a$ are:

\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{u(x)}\right) = e^{u(x)} u'(x), \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(b^{u(x)}\right) = b^{u(x)} \ln b \cdot u'(x)

\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln \abs{u(x)}\right) = \frac{u'(x)}{u(x)}, \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(\log_b \abs{u(x)}\right) = \frac{u'(x)}{u(x) \ln b}

The first two are the chain rule; the last two, its version on the domain of the composition, because the logarithm is not defined at $0$ .

Proof

$\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{u(x)}\right) = e^{u(x)} u'(x), \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(b^{u(x)}\right) = \resaltar{b^{u(x)} \ln b \cdot u'(x)}$
The outer functions $e^t$ and $b^t$ are differentiable on all of $\mathbb{R}$ , which contains the values of $u$ ; the chain rule gives the first two derivatives, at $x = a$ .
$\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln \abs{u(x)}\right) = \frac{1}{u(x)} \cdot u'(x) = \resaltar{\frac{u'(x)}{u(x)}}$
The function $\ln \abs{t}$ is defined on $B = \mathbb{R} \setminus \{0\}$ , which contains the interval with center $u(a)$ and radius $r = \abs{u(a)}$ , and it is differentiable at $u(a)$ , with derivative $\frac{1}{u(a)}$ . By the corollary on the chain rule on the domain of the composition, $a$ is an accumulation point of the domain of $\ln \abs{u(x)}$ , and the derivative at $x = a$ is:
$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log_b \abs{u(x)}\right) = \frac{1}{\ln b} \cdot \frac{u'(x)}{u(x)} = \resaltar{\frac{u'(x)}{u(x) \ln b}}$
If $u(a) > 0$ , the same corollary, with $B = \ioo{0}{\infty}$ and $r = u(a)$ , gives the derivative of $\ln u(x)$ , which is the same. For the base $b$ , one divides by the constant $\ln b$ , by the change of base.

Thus, the derivative of $2^{\sin x}$ is $2^{\sin x} \ln 2 \cdot \cos x$ , and that of $\ln\left(x^2 + 1\right)$ is $\frac{2x}{x^2 + 1}$ on all of $\mathbb{R}$ . The lesson on the chain rule differentiated $\ln(\cos x)$ on its domain, where $\cos x > 0$ , and obtained $-\tan x$ ; the absolute value extends the result: $\ln \abs{\cos x}$ is defined where $\cos x \neq 0$ , and its derivative is $-\tan x$ on that whole set, which is the domain of the tangent function.

In the rest of this lesson

Powers with real exponents
Powers with variable base and exponent
Logarithmic differentiation
Which is larger, pi to the e or e to the pi?
Solved problems
The derivative of x to the x
Logarithmic differentiation where the function is negative
A positive derivative near the number e
The two incomplete rules
The exponential tangent to the line y equals x
Proposed problems
The logarithm to base x
Two power towers
A product at one of its zeros
Rational powers of negative bases
Comparing a to the b with b to the a