AP Calculus AB and BC

Derivative of ln x, e^x, and a^x: Answers and Proofs

Q: What is the derivative of 2^x?

\frac{d}{dx} 2^x = 2^x \ln 2. It follows the general exponential rule \frac{d}{dx} a^x = a^x \ln a with a = 2. A frequent mistake is writing x \cdot 2^{x-1}, but that is the power rule, which applies only when the variable is in the base, not the exponent.

Q: Why does a^x differentiate to a^x ln a instead of just a^x?

Only base e gives a derivative with no extra factor, because e is defined so that \lim_{h \to 0} \frac{e^h - 1}{h} = 1. Rewriting a^x = e^{x \ln a} and applying the chain rule produces the inner derivative \ln a, so \frac{d}{dx} a^x = a^x \ln a. When a = e, \ln e = 1 and the factor disappears.

Q: What is the derivative of ln(x)?

\frac{d}{dx} \ln x = \frac{1}{x} for x > 0. If the logarithm wraps a more complicated expression, say \ln(x^2 + 4), the chain rule adds the derivative of the inside: \frac{1}{x^2 + 4} \cdot 2x. For negative inputs, \frac{d}{dx} \ln|x| = \frac{1}{x} holds for all x \ne 0.

The derivative of ln x is 1/x, the derivative of e^x is e^x itself, and the derivative of a^x is a^x times ln a (so d/dx of 2^x is 2^x ln 2). Each is proved below: e^x from the limit defining e, ln x by inverting x = e^y, and a^x by rewriting a^x as e^(x ln a).

Recognize the function before you pick a rule

Three rules hide behind functions that look nearly identical, and choosing the wrong one is the most common way to lose these points. Before you differentiate, ask two questions. First: is the variable in the exponent or in the base? Second: if the variable is in the exponent, is the base $e$ or some other constant $a$ ? Your answers pick the rule.

If the variable sits in the base with a constant exponent, like $x^5$ , that is the power rule from Topic 2.5, not an exponential rule at all. The functions on this page put the variable somewhere else: in the exponent, as in $e^x$ and $a^x$ , or inside a logarithm, as in $\ln x$ . CED Topic 2.7 supplies the rules for $e^x$ and $\ln x$ , and the $a^x$ result follows from a single rewrite that the later sections prove. The recognition habit worth building is to label the function by name (exponential, power, or logarithm) out loud before you write anything down, because the wrong label sends you to the wrong rule and no amount of careful algebra recovers from that.

Function	What to notice	Derivative
$e^x$	variable exponent, base $e$	$e^x$
$a^x$	variable exponent, constant base $a > 0$	$a^x \ln a$
$\ln x$	natural logarithm	$\frac{1}{x}$
$\log_a x$	logarithm, base $a$	$\frac{1}{x \ln a}$
$x^n$	constant exponent (power rule, not here)	$n x^{n-1}$

The $e^x$ and $\ln x$ rows are the Topic 2.7 rules to memorize; the $a^x$ and $\log_a x$ rows are not separate CED rules but derived results, following from those two through the $a^x = e^{x \ln a}$ and change-of-base rewrites proved in the later sections.

The a^x versus x^n trap

$2^x$ and $x^2$ are not the same kind of function. In $x^2$ the exponent is constant, so the power rule gives $2x$ . In $2^x$ the exponent is the variable, so the exponential rule gives $2^x \ln 2$ . Reading which slot the variable sits in, base or exponent, decides everything. Writing $\frac{d}{dx} 2^x = x \cdot 2^{x-1}$ applies the power rule to an exponential and is simply wrong.

Why e^x is its own derivative

Start from the limit definition of the derivative applied to $f(x) = e^x$ .

\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}

Use the exponent law $e^{x+h} = e^x \cdot e^h$ , then factor $e^x$ out of the numerator. That factor does not depend on $h$ , so it comes outside the limit.

= \lim_{h \to 0} \frac{e^x(e^h - 1)}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}

Everything now hinges on that last limit. The number $e$ is defined to be the one base for which $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ . For any other base the analogous limit is a different constant (it turns out to equal $\ln a$ , which the $a^x$ section confirms). Because the base here is $e$ , the limit is exactly $1$ , and the derivative collapses to $e^x$ itself.

\frac{d}{dx} e^x = e^x

That self-reproducing property, a function equal to its own rate of change, is why $e^x$ runs through exponential growth, differential equations, and Taylor series. Topic 2.7 lists it as a rule to memorize; the limit above is where the rule comes from.

Proving d/dx ln x = 1/x with inverse functions (Topic 3.3)

The natural logarithm is the inverse of $e^x$ : $\ln x$ and $e^x$ undo each other. Topic 3.3 says you can differentiate an inverse function using the chain rule together with the defining relationship, and $\ln x$ is the cleanest example. Set $y = \ln x$ . By the definition of the logarithm as an inverse, that is the same statement as this equation.

e^y = x

Differentiate both sides with respect to $x$ . The right side is just $x$ , whose derivative is $1$ . The left side is $e^y$ , where $y$ is a function of $x$ , so the chain rule applies: the derivative of $e^y$ is $e^y$ times $\frac{dy}{dx}$ .

e^y \cdot \frac{dy}{dx} = 1

Solve for $\frac{dy}{dx}$ , then replace $e^y$ with $x$ using $e^y = x$ from the setup.

\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}

So $\frac{d}{dx} \ln x = \frac{1}{x}$ , valid for $x > 0$ , the domain of $\ln x$ . A useful extension, $\frac{d}{dx} \ln|x| = \frac{1}{x}$ for all $x \ne 0$ , covers negative inputs and shows up whenever you antidifferentiate $\frac{1}{x}$ . This inverse-function move (write the inverse relation, differentiate implicitly, solve) is exactly the method Topic 3.3 asks you to reuse for inverse trig functions in Topic 3.4.

a^x = e^(x ln a): one rewrite handles every base

You do not need a separate rule for $a^x$ ; you need one algebra step. Any positive number $a$ can be written as $a = e^{\ln a}$ , because $e$ and $\ln$ are inverses. Raise both sides to the $x$ power and use the exponent law $(e^b)^x = e^{bx}$ .

a^x = \left(e^{\ln a}\right)^x = e^{x \ln a}

Now $a^x$ is an exponential with base $e$ and inner function $x \ln a$ , so the chain rule finishes it. Because $\ln a$ is a constant, the derivative of the inner function $x \ln a$ is just $\ln a$ .

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

This also settles the loose end from Section 2. Differentiating $a^x$ straight from the definition produces the factor $\lim_{h \to 0} \frac{a^h - 1}{h}$ , and the proof above shows that factor equals $\ln a$ . For base $e$ , $\ln e = 1$ , which is precisely why $e^x$ carries no extra constant.

Logarithms with other bases yield to the same trick through the change-of-base identity $\log_a x = \frac{\ln x}{\ln a}$ . Since $\frac{1}{\ln a}$ is a constant multiplier, differentiate using $\frac{d}{dx}\ln x = \frac{1}{x}$ .

\frac{d}{dx} \log_a x = \frac{1}{\ln a} \cdot \frac{d}{dx} \ln x = \frac{1}{x \ln a}

Where students lose points

Almost every error on these derivatives is one of a handful of predictable slips.

Treating $a^x$ like a power. $\frac{d}{dx} 2^x$ is $2^x \ln 2$ , not $x \cdot 2^{x-1}$ . The exponent is the variable, so the power rule does not apply.
Dropping the $\ln a$ factor. $\frac{d}{dx} 5^x = 5^x \ln 5$ ; writing just $5^x$ forgets the constant that every base other than $e$ carries.
Forgetting the chain rule inside a logarithm. $\frac{d}{dx} \ln(3x^2 + 1)$ is $\frac{1}{3x^2 + 1} \cdot 6x$ , not $\frac{1}{3x^2 + 1}$ . The $\frac{1}{x}$ rule is really $\frac{1}{\text{inside}}$ times the derivative of the inside.
Mixing up the two directions. The derivative of $\ln x$ is $\frac{1}{x}$ , but the antiderivative of $\frac{1}{x}$ is $\ln|x| + C$ ; keep which way you are going straight.
Assuming $e^{2x}$ differentiates to $e^{2x}$ . Only $e^x$ is its own derivative. $\frac{d}{dx} e^{2x} = 2e^{2x}$ , because the chain rule contributes the inner derivative $2$ .

One-line check

After differentiating any exponential, ask whether the base was $e$ . If not, a $\ln(\text{base})$ factor must appear. After differentiating any logarithm, ask whether the inside was more complicated than $x$ ; if so, a chain-rule factor must appear.

Worked examples

Worked example

Separate exponential, power, and natural-exponential terms

Differentiate $f(x) = 3^x + x^3 + e^x$ .

Classify each term before touching a rule. $3^x$ is exponential (variable in the exponent, base $3$ ), $x^3$ is a power (variable in the base, constant exponent), and $e^x$ is the natural exponential. Three look-alike terms, three different rules.
Differentiate $3^x$ with the exponential rule: the derivative of $a^x$ is $a^x \ln a$ . Here $a = 3$ , giving $3^x \ln 3$ .
Differentiate $x^3$ with the power rule: the derivative of $x^n$ is $n x^{n-1}$ . Here $n = 3$ , giving $3x^2$ .
Differentiate $e^x$ : it is its own derivative, $e^x$ .
Add the three results in order: $f'(x) = 3^x \ln 3 + 3x^2 + e^x$ .

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

Worked example

A logarithm that needs the chain rule

Differentiate $g(x) = \ln(x^2 + 4)$ .

Recognize the structure: a natural log wrapped around $x^2 + 4$ . The inside is more complicated than $x$ , so the $\frac{1}{x}$ rule needs a chain-rule factor. Name the inner function $u = x^2 + 4$ .
Apply the chain rule: if $y = \ln u$ , then $y' = \frac{1}{u} \cdot u'$ . The outer log gives $\frac{1}{x^2 + 4}$ , with the inside left intact.
Differentiate the inside: $u' = 2x$ .
Multiply the outer and inner pieces: $g'(x) = \frac{1}{x^2 + 4} \cdot 2x$ .
Combine into a single fraction: $g'(x) = \frac{2x}{x^2 + 4}$ .

$g'(x) = \frac{2x}{x^2 + 4}$

Worked example

Read a limit as a known derivative (Topic 2.7)

Evaluate $\lim_{h \to 0} \frac{2^h - 1}{h}$ .

This looks like an indeterminate $\frac{0}{0}$ limit, but Topic 2.7 flags a shortcut: sometimes a limit is exactly the definition of a derivative you already know. Compare it to the derivative definition $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ .
Match the pieces. If $f(x) = 2^x$ and $a = 0$ , then $f(a+h) = 2^{0+h} = 2^h$ and $f(a) = 2^0 = 1$ , so $\frac{f(a+h) - f(a)}{h} = \frac{2^h - 1}{h}$ . The limit is therefore $f'(0)$ for $f(x) = 2^x$ .
Differentiate with the exponential rule: $f'(x) = 2^x \ln 2$ .
Evaluate at $a = 0$ : $f'(0) = 2^0 \ln 2 = 1 \cdot \ln 2 = \ln 2$ .

$\lim_{h \to 0} \frac{2^h - 1}{h} = \ln 2$

Frequently asked questions

What is the derivative of 2^x?

$\frac{d}{dx} 2^x = 2^x \ln 2$ . It follows the general exponential rule $\frac{d}{dx} a^x = a^x \ln a$ with $a = 2$ . A frequent mistake is writing $x \cdot 2^{x-1}$ , but that is the power rule, which applies only when the variable is in the base, not the exponent.

Why does a^x differentiate to a^x ln a instead of just a^x?

Only base $e$ gives a derivative with no extra factor, because $e$ is defined so that $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ . Rewriting $a^x = e^{x \ln a}$ and applying the chain rule produces the inner derivative $\ln a$ , so $\frac{d}{dx} a^x = a^x \ln a$ . When $a = e$ , $\ln e = 1$ and the factor disappears.

What is the derivative of ln(x)?

$\frac{d}{dx} \ln x = \frac{1}{x}$ for $x > 0$ . If the logarithm wraps a more complicated expression, say $\ln(x^2 + 4)$ , the chain rule adds the derivative of the inside: $\frac{1}{x^2 + 4} \cdot 2x$ . For negative inputs, $\frac{d}{dx} \ln|x| = \frac{1}{x}$ holds for all $x \ne 0$ .