Differentiating aˣ and logₐ x

Any positive base is a power of e

Let a > 0. Then ln a is defined: it is the power to which e must be raised to give a. So $a = e^{\ln a}$. For example, ln 2 = 0.6931 to four decimal places, and $e^{\ln 2} = 2$.

Raise both sides to the power x: $a^{x} = \left(e^{\ln a}\right)^{x} = e^{x \ln a}$. So $2^{x} = e^{x \ln 2}$, which is about $e^{0.6931x}$. The base must be positive, because ln a has no value for $a \le 0$.

The chain rule on $e^{x \ln a}$

In $e^{x \ln a}$, the inner function is x ln a, where ln a is a constant, so its derivative is ln a. The outer function is eᵘ, which is its own derivative. The chain rule gives $e^{x \ln a} \times \ln a$.

Written back with base a, $e^{x \ln a}$ is $a^{x}$. So the derivative of $a^{x}$ is $a^{x} \ln a$.

Check $2^{x}$ at x = 3: the rule gives 8 ln 2 = 5.5452, and the chord from 3 to 3.001 has gradient 5.5471. Check $3^{x}$ at x = 2: the rule gives 9 ln 3 = 9.8875, and a chord with a step of 0.000001 gives 9.8875.

The multiplier is ln a

The gradient of $a^{x}$ is its height $a^{x}$ times the constant ln a. Chords at x = 0 give that constant as about 0.6931 for $2^{x}$ and 1.0986 for $3^{x}$, and those are ln 2 and ln 3.

ln 2 = 0.6931 is less than 1, so $2^{x}$ has a gradient less than its height at every point, and from the same start at (0, 1) it climbs more slowly than $e^{x}$. ln 3 = 1.0986 is more than 1, so $3^{x}$ climbs faster than its height. For $10^{x}$ the multiplier is ln 10 = 2.3026.

The gold curve $y = 2^{x}$ and the dashed curve $y = e^{x}$, both through (0, 1). The gold line is the tangent to $2^{x}$ there, with gradient ln 2 = 0.6931; the dashed line is the tangent to $e^{x}$, with gradient 1. At x = 2 the heights are 4 and 7.389.

Checks with other bases

With a = e, ln e = 1, and the rule gives $e^{x} \times 1 = e^{x}$: $e^{x}$ is its own derivative, as before.

With a = 1, ln 1 = 0, and the rule gives 0. That is right: $1^{x} = 1$ for every x, a horizontal line.

With 0 < a < 1, ln a is negative, and $a^{x}$ falls. For $\left(\frac{1}{2}\right)^{x}$ at x = 1, the gradient is $\left(\frac{1}{2}\right) \times \ln\left(\frac{1}{2}\right) = 0.5 \times \left(-0.6931\right) = -0.3466$.

$\log_{a} x$

By the change of base, $\log_{a} x = \ln x$ / ln a, for x > 0 and a positive base $a \ne 1$. Here $\frac{1}{{\ln a}}$ is a constant multiplier, so the derivative of $\log_{a} x$ is $\left(\frac{1}{\ln a}\right) \times \left(\frac{1}{x}\right) = \frac{1}{{x \ln a}}$.

For $\log_{2} x$ at x = 4 that is $\frac{1}{{4 \ln 2}} = 0.3607$, and a chord with a step of 0.000001 gives 0.3607. For $\log_{10} x$ at x = 10 it is $\frac{1}{{10 \ln 10}} = 0.04343$.

At x = 1, $\log_{2} x$ has gradient $\frac{1}{{\ln 2}} = 1.4427$, steeper than ln x, whose gradient there is 1. The graph of $\log_{2} x$ is the graph of ln x stretched vertically by the factor 1.4427.

A coefficient in the exponent

For $2^{3x} = e^{3x \ln 2}$, the inner function is 3x ln 2, with derivative 3 ln 2, so the derivative is $3 \ln 2 \times 2^{3x}$. At x = 1 that is 24 ln 2 = 16.636.

For $10^{-x}$, the inner derivative is −ln 10, so the derivative is $-\ln 10 \times 10^{-x}$: the function falls at 2.3026 times its own height.

The usual mistakes

Using the power rule, as $x \cdot 2^{x-1}$. The power rule needs the variable in the base and a fixed power; in $2^{x}$ the base is fixed and the variable is in the exponent.

Giving $2^{x}$ as its own derivative. Only base e has a gradient equal to its height; $2^{x}$ has ln 2 times its height.

Dividing by ln a instead of multiplying. The derivative of $a^{x}$ is $a^{x} \ln a$.

Giving $\frac{1}{x}$ as the derivative of $\log_{2} x$. That is the derivative of ln x; $\log_{2} x$ is ln x divided by ln 2, so its derivative is $\frac{1}{{x \ln 2}}$.

A tracer that halves every six hours

In the application below, a tracer’s activity is $800 \times 2^{- \frac{x}{6}}$. Its derivative needs the factor ln 2, and writing the same activity with base e shows where the ln 2 comes from.