Derivatives of Exponential and Logarithmic Functions

Exponential functions are exactly the functions whose rate of change is proportional to their own size (on an interval, $f'=kf$ forces $f(x)=Ce^{kx}$; the interval hypothesis matters, since on a disconnected domain each piece may carry its own $C$), which is why they govern compound interest, population growth, radioactive decay and Newtonian cooling; logarithms come along as their inverses. This topic pins down the number $e$ by exactly that property, derives the four core derivatives ($e^x$, $a^x$, $\ln x$, $\log_a x$), and then develops logarithmic differentiation, the technique that converts a monstrous product or quotient into a sum and is one of the two standard ways to handle a variable base with a variable exponent such as $x^x$ (the other being the rewrite $u^{v}=e^{v\ln u}$, valid when $u>0$). Everything here sits on top of the chain rule, which you will combine with every formula below.

Where the number $e$ comes from

Fix a base $a>0$ and let $f(x)=a^{x}$. From the limit definition of the derivative,

$$\begin{aligned} f'(x) &= \lim_{h\to 0}\frac{a^{x+h}-a^{x}}{h} \\ &= \lim_{h\to 0}\frac{a^{x}\left(a^{h}-1\right)}{h} \\ &= a^{x}\lim_{h\to 0}\frac{a^{h}-1}{h} = a^{x}\,f'(0). \end{aligned}$$

The factor $a^{x}$ comes out of the limit because it does not depend on $h$. So the derivative of every exponential function is a constant multiple of the function itself, and the constant is exactly the slope of its graph at $x=0$. Write $m(a)=\lim_{h\to 0}\dfrac{a^{h}-1}{h}=f'(0)$. One shows (this needs a separate argument, usually via the integral definition of $\ln$) that $m(a)$ exists for every $a>0$ and increases continuously in $a$, with $m(1)=0$ and $m(a)\to\infty$ as $a\to\infty$.

Definition. $e$ is the unique base for which that slope equals $1$:

$$\lim_{h\to 0}\frac{e^{h}-1}{h}=1, \qquad\text{equivalently}\qquad \frac{d}{dx}e^{x}=e^{x}.$$

Equivalent characterisations: $e=\lim_{h\to 0}(1+h)^{1/h}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=2.718281828\ldots$ Once the log derivative is available we get $m(a)=\ln a$ for free, so the mysterious constant $m(a)$ above was always a logarithm.

The four core derivatives

Natural exponential. $\dfrac{d}{dx}e^{x}=e^{x}$ for all real $x$ — this is the definition of $e$ applied to $f'(x)=a^{x}f'(0)$ with $a=e$.

General exponential ($a>0$). Because $a>0$, $\ln a$ exists and $a^{x}=\left(e^{\ln a}\right)^{x}=e^{x\ln a}$. The chain rule with inner function $u=x\ln a$, $u'=\ln a$, gives

$$\frac{d}{dx}a^{x}=e^{x\ln a}\cdot\ln a=a^{x}\ln a .$$

Sanity checks: $a=e$ gives $\ln e=1$; $a=1$ gives $\ln 1=0$, correct since $1^{x}$ is constant. The hypothesis $a>0$ cannot be dropped: for $a<0$ the power $a^{x}$ is not defined on any interval (only at scattered rationals), and for $a=0$ the function $0^{x}$ exists only on $(0,\infty)$, where it is identically $0$ while $\ln 0$ does not exist at all.

Natural logarithm. $\ln$ is the inverse of $\exp$; $\exp$ is differentiable, strictly increasing, maps the real line bijectively onto $(0,\infty)$, and has $\exp'=\exp\ne 0$ everywhere, so by the Inverse Function Theorem $\ln$ is differentiable on $(0,\infty)$. Put $y=\ln x$, so $e^{y}=x$ with $x>0$. Differentiating both sides with respect to $x$ (chain rule on the left):

$$e^{y}\frac{dy}{dx}=1 \quad\Longrightarrow\quad \frac{dy}{dx}=\frac{1}{e^{y}}=\frac{1}{x},\qquad x>0 .$$

Absolute-value version. For every $x\ne 0$,

$$\frac{d}{dx}\ln\left|x\right|=\frac{1}{x}.$$

Proof by cases: for $x>0$, $\ln|x|=\ln x$ and the derivative is $\frac1x$; for $x<0$, $\ln|x|=\ln(-x)$ and the chain rule gives $\frac{1}{-x}\cdot(-1)=\frac1x$. The point $x=0$ must be excluded — the function is not even defined there.

Other bases. For $a>0$, $a\ne 1$, and $x>0$, change of base gives $\log_a x=\dfrac{\ln x}{\ln a}$ with $\dfrac{1}{\ln a}$ a constant, so

$$\frac{d}{dx}\log_a x=\frac{1}{x\ln a}, \qquad \frac{d}{dx}\log_a\left|x\right|=\frac{1}{x\ln a}\ \ (x\ne 0).$$

Rule table (chain-rule form, $u=u(x)$ differentiable)

With absolute values the two log rows extend to $\dfrac{d}{dx}\ln\left|u\right|=\dfrac{u'}{u}$ wherever $u(x)\ne 0$ — a genuinely useful widening, since $u$ is allowed to be negative.

Logarithmic differentiation

Use it when $y$ is a product or quotient of three or more factors, when factors carry awkward powers or roots, or when the exponent itself contains $x$.

Method. Let $y=f(x)$ be differentiable with $f(x)\ne 0$ near the point of interest.

1. Write $\ln\left|y\right|=\ln\left|f(x)\right|$.
2. Break the right side up with the log laws: $\ln\left|AB\right|=\ln\left|A\right|+\ln\left|B\right|$, $\ln\left|A/B\right|=\ln\left|A\right|-\ln\left|B\right|$, and $\ln\left|A^{p}\right|=p\ln\left|A\right|$ (integer $p$; for a general real exponent you need $A>0$).
3. Differentiate both sides in $x$. The chain rule turns the left side into $\dfrac{y'}{y}$.
4. Multiply through by $y$ and substitute the original expression back in.

Hypotheses that matter. Step 1 requires $f(x)\ne 0$: at a zero of $f$ the method says nothing and you must differentiate directly. The log laws split products, quotients and powers only — there is no law for $\ln(A+B)$, so logarithmic differentiation does not help with sums.

Worked Example 1 — a heavy quotient. Differentiate $y=\dfrac{x^{3}\sqrt{2x+1}}{(3x-2)^{4}}$ for $x>\frac{2}{3}$ (where all three factors are positive).

$$\ln y=3\ln x+\frac{1}{2}\ln(2x+1)-4\ln(3x-2).$$

Differentiate term by term, each with the chain rule $\left(\ln u\right)'=u'/u$:

$$\begin{aligned} \frac{y'}{y} &= \frac{3}{x}+\frac{1}{2}\cdot\frac{2}{2x+1}-4\cdot\frac{3}{3x-2} \\ &= \frac{3}{x}+\frac{1}{2x+1}-\frac{12}{3x-2}. \end{aligned}$$

Multiply by $y$:

$$y'=\frac{x^{3}\sqrt{2x+1}}{(3x-2)^{4}}\left(\frac{3}{x}+\frac{1}{2x+1}-\frac{12}{3x-2}\right).$$

Check at $x=1$. Then $y=\sqrt{3}$ and the bracket is $3+\frac{1}{3}-12=-\frac{26}{3}$, so $y'(1)=-\frac{26\sqrt{3}}{3}\approx-15.0111$. A centred difference quotient with step $10^{-6}$ gives $-15.0111$.

Worked Example 2 — variable base and variable exponent. Differentiate $y=x^{\sin x}$ for $x>0$.

The power rule is illegal (the exponent is not constant) and the $a^{x}$ rule is illegal (the base is not constant). Two equivalent routes.

Route A (rewrite as base $e$). Since $x>0$, $x^{\sin x}=e^{\sin x\ln x}$, so by the chain rule with the product rule inside,

$$y'=e^{\sin x\ln x}\left(\cos x\ln x+\frac{\sin x}{x}\right)=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right).$$

Route B (take logs). $\ln y=\sin x\ln x$, so $\dfrac{y'}{y}=\cos x\ln x+\dfrac{\sin x}{x}$, and multiplying by $y$ gives the same thing.

Check at $x=\frac{\pi}{2}$. There $y=\left(\frac{\pi}{2}\right)^{1}=\frac{\pi}{2}$, $\cos\frac{\pi}{2}=0$, $\sin\frac{\pi}{2}=1$, so $y'=\frac{\pi}{2}\left(0+\frac{2}{\pi}\right)=1$; the numerical derivative is $1.0000$.

Why the power rule cannot touch $x^{x}$

$\dfrac{d}{dx}x^{n}=nx^{\,n-1}$ is proved under the standing hypothesis that $n$ is a constant, and $\dfrac{d}{dx}a^{x}=a^{x}\ln a$ under the hypothesis that $a$ is a constant. In $x^{x}$ both move at once, so neither theorem applies. The correct general statement, for $u(x)>0$, comes from $u^{v}=e^{v\ln u}$:

$$\frac{d}{dx}u^{v}=u^{v}\left(v'\ln u+\frac{v\,u'}{u}\right).$$

The two familiar rules are its degenerate cases: if $v$ is constant then $v'=0$ and the formula collapses to $v\,u^{\,v-1}u'$; if $u$ is constant then $u'=0$ and it collapses to $u^{v}\ln u\cdot v'$. For $u=v=x$ neither term dies:

$$\frac{d}{dx}x^{x}=x^{x}\left(1\cdot\ln x+\frac{x\cdot 1}{x}\right)=x^{x}\left(\ln x+1\right),\qquad x>0 .$$

At $x=2$ the true slope is $4(\ln 2+1)\approx 6.7726$; the two popular wrong answers give $x\cdot x^{\,x-1}=x^{x}=4$ and $x^{x}\ln x\approx 2.7726$. Both are badly wrong, and the numerical derivative $6.7726$ settles it.

Common mistakes

Two identities students overuse: $e^{\ln x}=x$ holds only for $x>0$, while $\ln\left(e^{x}\right)=x$ holds for every real $x$.

Fast sanity checks

• Any derivative of $e^{\square}$ must still contain $e^{\square}$ as a factor — exponentials never disappear under differentiation.
• Any derivative of $\ln(\text{something})$ must carry that something in a denominator, since $\left(\ln u\right)'=u'/u$. A leftover $\ln$ outside that fraction usually means you differentiated the wrong layer — the honest exceptions are nested logs, where $\dfrac{d}{dx}\ln\left(\ln x\right)=\dfrac{1}{x\ln x}$ for $x>1$, and powers of logs, where $\dfrac{d}{dx}\left(\ln x\right)^{2}=\dfrac{2\ln x}{x}$; in both the surviving $\ln$ is correct.
• If your answer for a variable-base-variable-exponent problem has only one term inside the bracket, you probably dropped either the $\ln u$ term or the $u'/u$ term.