AP Calculus AB and BC

Derivative of e^x: Answer, Proof, and Mistakes

Q: Is the derivative of e^x really just e^x?

Yes. e^{x} is the one function, up to constant multiples, that equals its own derivative. It falls out of the limit definition: \frac{d}{dx}e^{x}=e^{x}\lim_{h\to 0}\frac{e^{h}-1}{h}, and e is defined precisely so that limit is 1.

Q: What is the derivative of e^(2x) or e^(kx)?

Use the chain rule: \frac{d}{dx}e^{kx}=k\,e^{kx}, so \frac{d}{dx}e^{2x}=2e^{2x}. The outside e^{kx} stays put, and you multiply by the derivative of the exponent.

Q: Why can't I use the power rule on e^x?

The power rule \frac{d}{dx}x^{n}=nx^{n-1} works only when the variable is the base and the exponent is constant. In e^{x} the base is the constant e and the variable is in the exponent, so it is exponential, not a power. Writing x\,e^{x-1} is wrong.

Q: How is the derivative of e^x different from a^x?

For any base, \frac{d}{dx}a^{x}=a^{x}\ln a. When a=e, \ln e=1, so the factor drops and \frac{d}{dx}e^{x}=e^{x}. That missing \ln a is exactly what makes e the natural base.

The derivative of e^x is e^x. The natural exponential function equals its own derivative, the only function (apart from constant multiples) that does, so d/dx of e^x = e^x. This follows straight from the limit that defines the number e. Do not use the power rule: e^x is not x times e^(x-1).

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

Why e^x equals its own derivative

Start from the limit definition of the derivative and factor. Because $e^{x+h}=e^{x}\cdot e^{h}$ , the fixed factor $e^{x}$ pulls straight out of the limit, and everything that depends on $h$ collects into one small piece.

\frac{d}{dx}e^{x}=\lim_{h \to 0}\frac{e^{x+h}-e^{x}}{h}=\lim_{h \to 0}\frac{e^{x}\left(e^{h}-1\right)}{h}=e^{x}\lim_{h \to 0}\frac{e^{h}-1}{h}

So the whole derivative rides on one limit: $\lim_{h \to 0}\frac{e^{h}-1}{h}$ . The number $e$ is defined to be the exact base that makes this limit equal to $1$ (equivalently, the base whose graph has slope $1$ at $x=0$ ). Substituting that value finishes the proof.

\lim_{h \to 0}\frac{e^{h}-1}{h}=1 \quad\Longrightarrow\quad \frac{d}{dx}e^{x}=e^{x}\cdot 1=e^{x}

Because $\frac{d}{dx}e^{x}=e^{x}$ , differentiating never changes it, and neither does antidifferentiating. It is the only function, apart from constant multiples $Ce^{x}$ , with that property: any $y$ satisfying $y'=y$ must equal $Ce^{x}$ .

This is what separates $e^{x}$ from its look-alikes. In $x^{e}$ the variable is the base and the exponent is the constant $e \approx 2.718$ , so the power rule applies and gives $e\,x^{e-1}$ . In a general exponential $a^{x}$ the derivative is $a^{x}\ln a$ ; $e^{x}$ is simply the case $a=e$ , where $\ln e=1$ and the extra factor disappears.

Function	Type	Rule that applies	Derivative
$e^{x}$	exponential, base $e$	equals its own derivative	$e^{x}$
$a^{x}$	exponential, base $a>0$	multiply by $\ln a$	$a^{x}\ln a$
$x^{e}$	power, constant exponent	power rule	$e\,x^{e-1}$

Where e^x shows up on the AP exam

The rule first appears in CED Topic 2.7, Derivatives of cos x, sin x, e^x, and ln x, which names $e^{x}$ outright. That topic sits in Unit 2 (Differentiation: Definition and Fundamental Properties), and Unit 2 carries 10 to 15 percent of the AB exam weighting and 5 to 10 percent on BC.

Most exam appearances are composites, not the bare $e^{x}$ . Once a function like $e^{3x}$ or $e^{-x^{2}}$ shows up, you layer the chain rule from CED Topic 3.1 (Unit 3, Differentiation: Composite, Implicit, and Inverse Functions) on top of the base rule.

\frac{d}{dx}\,e^{g(x)}=e^{g(x)}\cdot g'(x)

Composite derivatives such as $e^{kx}$ , $e^{-x^{2}}$ , and $e^{\sin x}$ , handled with the chain rule (Topic 3.1).
Product and quotient combinations like $x^{2}e^{x}$ or $\frac{e^{x}}{x}$ , where $e^{x}$ is one factor (Topics 2.8 and 2.9).
Exponential growth and decay solutions $y=Ce^{kt}$ from separable differential equations (Unit 7).
Tangent-line, linearization, and L'Hopital's Rule problems where you evaluate $e^{x}$ and its derivative at a point (Unit 4).

The self-replicating behavior runs backward too: $\int e^{x}\,dx=e^{x}+C$ (CED Topic 6.8). The natural exponential is its own antiderivative as well as its own derivative, which is why $e^{x}$ survives every round of calculus you do to it.

Common mistakes

Almost every lost point on this rule traces back to one of four specific errors. Each has a clean fix once you name what went wrong.

Using the power rule. The exponent is the variable, so $e^{x}$ does not become $x\,e^{x-1}$ . The power rule is only for a constant exponent on a variable base, like $x^{e}$ .
Dropping the inside derivative. $\frac{d}{dx}e^{3x}=3e^{3x}$ , not $e^{3x}$ ; the chain rule factor of $3$ is part of the answer.
Losing the sign on $e^{-x}$ . The inside derivative is $-1$ , so $\frac{d}{dx}e^{-x}=-e^{-x}$ .
Over-generalizing to other bases. $\frac{d}{dx}2^{x}=2^{x}\ln 2$ , not $2^{x}$ . Only base $e$ drops the $\ln a$ factor, because $\ln e=1$ .

Quick self-check: if your answer to $\frac{d}{dx}e^{x}$ has an $x$ in it, you used the power rule by mistake. The correct answer is exactly $e^{x}$ , with no coefficient and no shifted exponent.

Chain rule composites, worked

The recipe for any $e^{g(x)}$ is the same: copy $e^{g(x)}$ down unchanged, then multiply by the derivative of the exponent. Two standard exam composites make the pattern concrete.

\frac{d}{dx}\,e^{3x}=e^{3x}\cdot\frac{d}{dx}(3x)=e^{3x}\cdot 3=3e^{3x}

The second example has a polynomial inside, so the multiplier is no longer a constant. It appears constantly in normal-distribution and other statistics setups.

\frac{d}{dx}\,e^{-x^{2}}=e^{-x^{2}}\cdot\frac{d}{dx}\left(-x^{2}\right)=e^{-x^{2}}\cdot(-2x)=-2x\,e^{-x^{2}}

One more pattern worth memorizing: for $e^{\sin x}$ the outside stays $e^{\sin x}$ and you multiply by $\cos x$ , giving $\frac{d}{dx}e^{\sin x}=\cos x \cdot e^{\sin x}$ . Same recipe, trig inside.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

Is the derivative of e^x really just e^x?

Yes. $e^{x}$ is the one function, up to constant multiples, that equals its own derivative. It falls out of the limit definition: $\frac{d}{dx}e^{x}=e^{x}\lim_{h\to 0}\frac{e^{h}-1}{h}$ , and $e$ is defined precisely so that limit is $1$ .

What is the derivative of e^(2x) or e^(kx)?

Use the chain rule: $\frac{d}{dx}e^{kx}=k\,e^{kx}$ , so $\frac{d}{dx}e^{2x}=2e^{2x}$ . The outside $e^{kx}$ stays put, and you multiply by the derivative of the exponent.

Why can't I use the power rule on e^x?

The power rule $\frac{d}{dx}x^{n}=nx^{n-1}$ works only when the variable is the base and the exponent is constant. In $e^{x}$ the base is the constant $e$ and the variable is in the exponent, so it is exponential, not a power. Writing $x\,e^{x-1}$ is wrong.

How is the derivative of e^x different from a^x?

For any base, $\frac{d}{dx}a^{x}=a^{x}\ln a$ . When $a=e$ , $\ln e=1$ , so the factor drops and $\frac{d}{dx}e^{x}=e^{x}$ . That missing $\ln a$ is exactly what makes $e$ the natural base.