Derivatives of Exponential and Logarithmic Functions

By the end of this lesson, you will be able to differentiate $e^x$, $\ln x$, $a^x$, and $\log_a x$, understand why $e$ is the natural base for calculus, and apply logarithmic differentiation.

The Derivative of $e^x$

$$\frac{d}{dx}[e^x] = e^x$$

$e^x$ is the only function (up to scalar multiples) that is its own derivative. This is the defining property of $e$ in calculus and why it's called the "natural" base.

Proof sketch:

$$\frac{d}{dx}[e^x] = \lim_{h\to 0}\frac{e^{x+h}-e^x}{h} = e^x\lim_{h\to 0}\frac{e^h - 1}{h} = e^x \cdot 1 = e^x$$

where $\lim_{h\to 0}(e^h-1)/h = 1$ follows from the definition of $e$.

The Derivative of $\ln x$

$$\frac{d}{dx}[\ln x] = \frac{1}{x} \quad (x > 0)$$

Proof: Let $y = \ln x$, so $e^y = x$. Differentiate implicitly: $e^y \cdot y' = 1$, so $y' = 1/e^y = 1/x$.

General Bases

$$\frac{d}{dx}[a^x] = a^x\ln a \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x\ln a}$$

Why: $a^x = e^{x\ln a}$; use the chain rule: $(e^{x\ln a})' = e^{x\ln a}\cdot\ln a = a^x\ln a$.

When $a = e$: $\ln e = 1$, so these reduce to the natural formulas. ✓

Example 1 — Direct Application

$f(x) = 3e^x - 4\ln x + x^2$

$f'(x) = 3e^x - \dfrac{4}{x} + 2x$

Example 2 — Product with Exponential

$g(x) = x^3 e^x$

$g'(x) = 3x^2 e^x + x^3 e^x = x^2 e^x(3 + x)$

Logarithmic Differentiation

For products/quotients of many terms, or functions of the form $f(x)^{g(x)}$:

Take $\ln$ of both sides, differentiate implicitly, solve for $y'$.

Example 3 — $x^x$

$y = x^x$ (not a power rule application — both base and exponent vary).

$\ln y = x\ln x$

Differentiate: $\dfrac{y'}{y} = \ln x + x\cdot\dfrac{1}{x} = \ln x + 1$

$y' = y(\ln x + 1) = x^x(\ln x + 1)$

Example 4 — Complex Product/Quotient

$y = \dfrac{x^3(x-1)^2}{\sqrt{x^2+1}}$

$\ln y = 3\ln x + 2\ln(x-1) - \frac{1}{2}\ln(x^2+1)$

$\dfrac{y'}{y} = \dfrac{3}{x} + \dfrac{2}{x-1} - \dfrac{x}{x^2+1}$

$y' = y\left(\dfrac{3}{x} + \dfrac{2}{x-1} - \dfrac{x}{x^2+1}\right)$