Derivative of e^x: The Function Equal to Its Own Derivative

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | FAQ | How This Fits

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The Principle

The move: Differentiate $e^x$ with respect to $x$ in one step: its derivative is $e^x$ again.

The invariant: Differentiating $e^x$ leaves the functional form unchanged: the exact derivative is $e^x$ again.

Pattern: $\frac{d}{dx} e^x \quad \longrightarrow \quad e^x$

Legal ✓	Illegal ✗
$\frac{d}{dx} e^x \to e^x$	$\frac{d}{dx} e^{2x} \to e^{2x}$

The illegal move treats $e^{2x}$ as though it were the direct $e^x$ pattern. The correct derivative is found by chain rule:

$\frac{d}{dx} e^{2x} = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x}$

The mistake is not a failure of the canonical condition. It is a wrong rule choice for the local structure.

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Conditions of Applicability

Condition: always applies

Before applying, check: are you taking the derivative of the exact local pattern $\frac{d}{dx} e^x$ as a direct step? If yes, apply derivativeExpRule directly. If the exponent is another expression, chain rule governs the step. If the term sits inside a sum, product, or constant multiple, that larger rule governs the whole derivative while derivativeExpRule is used locally on the $e^x$ term.

• The canonical condition is still “always applies.” The real decision point here is structural recognition: are you looking at the direct $e^x$ pattern, or at a larger structure that calls another rule first?
• If the exponent is a differentiable function $g(x)$ rather than the bare variable $x$, chain rule supplies the extra factor: $\frac{d}{dx} e^{g(x)} = g'(x)e^{g(x)}$.
• If the base is not $e$ but a general constant $a$, use the general derivative rule for $a^x$ instead: $\frac{d}{dx} a^x = a^x \ln a$.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: treat every exponential with base $e$ as the direct $e^x$ pattern and write $\frac{d}{dx} e^{g(x)} = e^{g(x)}$ — dropping the chain factor $g'(x)$ and misidentifying which rule governs the step.

Debug: inspect the local structure before writing the derivative. If the step is exactly $\frac{d}{dx} e^x$, use derivativeExpRule directly. If the exponent is anything other than the bare variable, write the chain-rule factor explicitly: $\frac{d}{dx} e^{g(x)} = g'(x)e^{g(x)}$.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What is special about the number $e$ that makes $e^x$ its own derivative — and would any other base $a$ share this property?
• Why does the derivative of $e^x$ equal $e^x$ exactly, rather than a constant multiple of $e^x$?

For the Principle

• How do you decide whether derivativeExpRule applies directly to the current local step, or whether chain rule (or product rule) governs the larger expression first?
• What changes in the derivative computation when the exponent is $2x$ instead of $x$?

Between Principles

• How does derivativeExpRule relate to the chain rule when the exponent is a differentiable function $g(x)$ — what extra factor appears, and where does it come from?

Generate an Example

• Describe a step in a derivative computation where applying derivativeExpRule without first inspecting the exponent would produce a wrong answer.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the derivative of e^x in one sentence: _____The derivative of e^x with respect to x is e^x — the natural exponential is its own derivative.

Write the canonical equation: _____$\frac{d}{dx} e^x = e^x$

State the canonical condition: _____always applies

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $\frac{d}{dx}(e^x + e^{2x})$, differentiate and track which term uses derivativeExpRule directly versus which term requires chain rule.

Step	Expression	Operation
0	$\frac{d}{dx}(e^x + e^{2x})$	—
1	$\frac{d}{dx}(e^x) + \frac{d}{dx}(e^{2x})$	Sum rule: distribute the derivative over addition
2	$e^x + \frac{d}{dx}(e^{2x})$	derivativeExpRule applies directly to the first term because it is exactly $e^x$
3	$e^x + e^{2x}\cdot\frac{d}{dx}(2x)$	Chain rule on the second term: outer $e^u$, inner $u=2x$
4	$e^x + e^{2x}\cdot 2$	Differentiate the inner function
5	$e^x + 2e^{2x}$	Simplify

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Drills

Forward step — differentiate using derivativeExpRule

Apply the rule to differentiate this expression.

$\frac{d}{dx}\, e^x$

Reveal

This is exactly the direct $\frac{d}{dx} e^x$ pattern, so derivativeExpRule applies directly:

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

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Apply the rule (with constant multiple rule) to differentiate this expression.

$\frac{d}{dx}\,(7e^x)$

Reveal

Constant multiple rule first; inside that step, this is exactly the direct $\frac{d}{dx} e^x$ pattern:

$\frac{d}{dx}(7e^x) = 7\,\frac{d}{dx}(e^x) = 7e^x$

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Evaluate: is the following derivation correct?

$\frac{d}{dx}\,e^{2x} = e^{2x}$

Reveal

No. This is not the direct $\frac{d}{dx} e^x$ pattern, so chain rule governs the step:

$\frac{d}{dx}\,e^{2x} = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x}$

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Does derivativeExpRule apply directly to the whole step below? If not, name the governing rule and complete the derivative.

$\frac{d}{dx}(xe^x)$

Reveal

No. The whole expression is a product, so product rule governs the derivative, with derivativeExpRule used on the $e^x$ factor inside that computation:

$\frac{d}{dx}(xe^x) = x\,\frac{d}{dx}(e^x) + e^x\,\frac{d}{dx}(x) = xe^x + e^x = e^x(x+1)$

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Differentiate $e^x + 3x^4$.

Reveal

$\frac{d}{dx}(e^x + 3x^4) = \frac{d}{dx}(e^x) + 3\,\frac{d}{dx}(x^4) = e^x + 3(4x^3) = e^x + 12x^3$

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Differentiate $-2e^x + x^3$.

Reveal

$\frac{d}{dx}(-2e^x + x^3) = -2\,\frac{d}{dx}(e^x) + \frac{d}{dx}(x^3) = -2e^x + 3x^2$

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Action label — identify which rule was applied

What was done between these two steps?

$\frac{d}{dx}(e^x + x^5) \quad\longrightarrow\quad e^x + 5x^4$

Reveal

Sum rule split the derivative across the two terms; then derivativeExpRule was used on $e^x$ and the power rule was used on $x^5$.

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What was done between these two steps?

$\frac{d}{dx}(8e^x) \quad\longrightarrow\quad 8e^x$

Reveal

Constant multiple rule preserved the factor $8$, and derivativeExpRule differentiated the $e^x$ term: $\frac{d}{dx}(8e^x) = 8\,\frac{d}{dx}(e^x) = 8e^x$.

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A student writes: $\frac{d}{dx}\,2^x = 2^x$. What error was made?

Reveal

derivativeExpRule applies only to base $e$, not to a general base $a$. For $a^x$ the correct rule introduces a logarithmic factor:

$\frac{d}{dx}\,2^x = 2^x \ln 2$

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In the computation below, name the rule applied to each term:

$\frac{d}{dx}(e^x - 3) \quad\longrightarrow\quad e^x - 0 \quad\longrightarrow\quad e^x$

Reveal

• $e^x$ term: derivativeExpRule — $\frac{d}{dx}(e^x) = e^x$.
• $-3$ term: derivative constant rule — $\frac{d}{dx}(-3) = 0$.

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Rule attribution — locate derivativeExpRule in a larger computation

In computing $\frac{d}{dx}(e^x + \ln x)$, which term uses derivativeExpRule and which uses a different rule?

Reveal

• $e^x$ term: derivativeExpRule — $\frac{d}{dx}(e^x) = e^x$.
• $\ln x$ term: derivative of ln(x) rule — $\frac{d}{dx}(\ln x) = \frac{1}{x}$.

So $\frac{d}{dx}(e^x + \ln x) = e^x + \frac{1}{x}$.

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For each expression below, state whether derivativeExpRule applies directly or another rule must be invoked first:

(a) $e^x$ — (b) $e^{x^2}$ — (c) $e^x \cdot x$

Reveal

(a) Direct — this is exactly the local pattern $\frac{d}{dx} e^x$, so derivativeExpRule applies immediately.

(b) Chain rule first — the exponent is $x^2$, so the derivative is $\frac{d}{dx}\,e^{x^2} = 2x\,e^{x^2}$.

(c) Product rule first — this is a product of $e^x$ and $x$. derivativeExpRule is used on the $e^x$ factor within the product-rule computation, but the governing rule for the whole step is product rule: $\frac{d}{dx}(e^x \cdot x) = e^x \cdot x + e^x \cdot 1 = e^x(x+1)$.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Find the derivative of $f(x) = 2e^x - 3x^2 + 5$.

Full solution

Step	Expression	Move
0	$2e^x - 3x^2 + 5$	—
1	$\frac{d}{dx}(2e^x) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(5)$	Sum/difference rule: distribute derivative over each term
2	$2\,\frac{d}{dx}(e^x) - 3\,\frac{d}{dx}(x^2) + \frac{d}{dx}(5)$	Constant multiple rule: pull coefficients out
3	$2e^x - 3\,\frac{d}{dx}(x^2) + 0$	derivativeExpRule on $e^x$; constant rule on $+5$ gives $0$
4	$2e^x - 3(2x)$	Power rule: $\frac{d}{dx}(x^2) = 2x$
5	$2e^x - 6x$	Arithmetic: $3 \times 2 = 6$

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FAQ

What is the derivative of $e^x$?

The derivative of $e^x$ with respect to $x$ is $e^x$. The natural exponential is its own derivative.

What is the canonical condition for derivativeExpRule?

The canonical condition is always applies. The real challenge is move selection: recognizing when the local step is exactly $\frac{d}{dx} e^x$ and when a neighboring rule governs the whole derivative.

Does the derivative of $e^x$ change if there is a coefficient in front, like $5e^x$?

No. Apply the constant multiple rule first: $\frac{d}{dx}(5e^x) = 5\,\frac{d}{dx}(e^x) = 5e^x$. The coefficient carries through unchanged.

What if the exponent is not just $x$ — for example $e^{2x}$ or $e^{x^2}$?

Then the direct one-step pattern is no longer the whole story, and chain rule governs the derivative: $\frac{d}{dx}\,e^{g(x)} = g'(x)e^{g(x)}$. For example, $\frac{d}{dx}\,e^{2x} = 2e^{2x}$ and $\frac{d}{dx}\,e^{x^2} = 2x\,e^{x^2}$.

How is $\frac{d}{dx} e^x$ different from $\frac{d}{dx} a^x$?

For general base $a \neq e$, the derivative is $a^x \ln a$. The $e^x$ rule is the special case where $\ln e = 1$, so the logarithmic factor disappears.

Can I use derivativeExpRule on $x^e$?

No — $x^e$ is a power function (constant exponent $e$, variable base), not an exponential function. Apply the power rule instead: $\frac{d}{dx}\,x^e = e\,x^{e-1}$.

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How This Fits in Unisium

Within the calculus subdomain, derivativeExpRule is one of the first exponential derivative rules you encounter. The main challenge is not memorizing a hidden applicability guard; it is recognizing local structure correctly. The drills above train you to separate the direct $e^x$ step from neighboring cases such as $e^{g(x)}$, $ce^x$, and products containing $e^x$, so that you choose the governing rule correctly before differentiating.

Explore further:

• Calculus Subdomain Map — Return to the calculus hub to see where the natural exponential rule sits in the derivative family
• Chain Rule — The direct $e^x$ rule becomes a chain-rule problem as soon as the exponent is any nontrivial function $g(x)$
• Derivative of ln(x) — The paired natural logarithm rule sits next to $e^x$ because the two functions are inverse partners on their real domains
• Principle Structures — Where derivativeExpRule fits in the calculus derivative taxonomy
• Elaborative Encoding — Build deep understanding of why $e^x$ is uniquely self-replicating under differentiation
• Retrieval Practice — Make the pattern and condition instantly accessible from memory

Ready to master the derivative of $e^x$? Check access and join the Unisium waitlist or explore the full learning framework in Masterful Learning.

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