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 $a=e^{\ln a}$a = e^(ln a). For example, ln 2 = 0.6931 to four decimal places, and $e^{\ln 2}=2$e^(ln 2) = 2.

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

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

In $e^{x\ln a}$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 $e^{x\ln a}\times \ln a$e^(x ln a) × ln a.

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

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

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

The gold curve $y=2^x$y = 2ˣ and the dashed curve $y=e^x$y = eˣ, both through (0, 1). The gold line is the tangent to $2^x$2ˣ there, with gradient ln 2 = 0.6931; the dashed line is the tangent to $e^x$eˣ, 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 $e^x\times 1=e^x$eˣ × 1 = eˣ: $e^x$eˣ is its own derivative, as before.

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

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

${\log }_{a}x$logₐ x

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

For ${\log }_{2}x$log₂ x at x = 4 that is $\frac{1}{4\ln 2}=0.3607$1/(4 ln 2) = 0.3607, and a chord with a step of 0.000001 gives 0.3607. For ${\log }_{10}x$log₁₀ x at x = 10 it is $\frac{1}{10\ln 10}=0.04343$1/(10 ln 10) = 0.04343.

At x = 1, ${\log }_{2}x$log₂ x has gradient $\frac{1}{\ln 2}=1.4427$1/(ln 2) = 1.4427, steeper than ln x, whose gradient there is 1. The graph of ${\log }_{2}x$log₂ 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}$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}$3 ln 2 × 2^(3x). At x = 1 that is 24 ln 2 = 16.636.

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

The usual mistakes

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

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

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

Giving $\frac{1}{x}$1/x as the derivative of ${\log }_{2}x$log₂ x. That is the derivative of ln x; ${\log }_{2}x$log₂ x is ln x divided by ln 2, so its derivative is $\frac{1}{x\ln 2}$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}}$800 × 2^(−x/6). Its derivative needs the factor ln 2, and writing the same activity with base e shows where the ln 2 comes from.