Mathematics · Calculus

Derivative of the Exponential Function

Q: How do you differentiate $a^{x}$, for example $2^{x}$?

Rewrite the base: $a^{x} = e^{(x\cdot{}\ln{} a)}$, since $e^{(\ln{} a)} = a$. Now the chain rule applies with inner derivative ln a, so $(a^{x})$′ = ln(a)·a^x. For $2^{x}$ this means $(2^{x})$′ = ln(2)·2^x ≈ 0.693·2^x, for $10^{x}$ correspondingly $\ln{}(10)\cdot{}10^{x} \approx{} 2.303\cdot{}10^{x}$. The factor ln a is the growth rate: for $a > e$ it exceeds 1, for $a < e$ it is smaller. The formula also shows why e is the "natural" base: only for $a = e$ does $\ln{} a = 1$ and the prefactor disappear. Exams like to use $2^{x}$ as a trap, because many reflexively write $2^{x}$ or $x\cdot{}2^{(x-1)}$; both are wrong.

Q: How do I differentiate products like $x\cdot{}e^{x}$ or $x^{2}\cdot{}e^{(-x)}$?

Here you combine product rule and chain rule. For f(x) = x·eˣ, f′(x) = 1·eˣ + x·eˣ = (1 + x)·eˣ. For $g(x) = x^{2}\cdot{}e^{(-x)}$ the product rule gives g′$(x) = 2x\cdot{}e^{(-x)} + x^{2}\cdot{}(-1)\cdot{}e^{(-x)} = (2x - x^{2})\cdot{}e^{(-x)}$; the inner derivative −1 comes from the chain rule. Factoring out the exponential at the end is practical: since eˣ or $e^{(-x)}$ is never zero, you read the zeros of the derivative from the polynomial factor alone, here $2x - x^{2} = x(2 - x)$ with candidates $x = 0$ and $x = 2$. Exactly this pattern, polynomial times exponential, is the most common function type in final-exam curve analysis.

Q: Why does the exponential function describe growth and decay processes?

Because its derivative is proportional to itself: $f(x) = c\cdot{}e^{(kx)}$ satisfies f′ = k·f. Exactly this characterizes natural processes where the rate of change depends on the current amount: the more bacteria there are, the more are added per hour; the more radioactive nuclei exist, the more decay per second. Positive k means growth, negative k decay. The exponential family is the only class of functions with this property, which is why it solves the differential equation f′ = k·f. In applications you determine k from data such as doubling time $(k =\frac{ \ln{} 2}{T_{D}})$ or half-life $(k = -\frac{\ln{} 2}{T_{H}})$. Thus the differentiation rule becomes a direct modelling tool for biology, physics and finance.

The exponential function is its own derivative; for $e^{(kx)}$ the chain rule adds the factor k.

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Formula

\frac{d}{d x} e^{x} = e^{x}, \frac{d}{d x} e^{k x} = k \cdot e^{k x}

Dimensionless (calculus)

Variables & units

Variables & units
Symbol	Meaning	Unit
$e$	Euler number $(\approx{} 2.71828)$	dimensionless
$x$	Independent variable in the exponent	dimensionless
$k$	Constant factor in the exponent (growth rate)	dimensionless

LaTeX

\frac{d}{dx}\, e^{x} = e^{x}, \quad \frac{d}{dx}\, e^{kx} = k \cdot e^{kx}

QUANTA STUDYquanta-study.de

Diagram: the graph of e to the x with a tangent at one point; the tangent slope equals the function value there. — The exponential is its own derivative: the tangent slope at each point equals the function value.

Derivation & background

Up to constant multiples, the exponential function is the only function that equals its own derivative. Exactly this property distinguishes the base e ≈ 2.71828 (Euler, 1748). In general $(e^{kx})' = ke^{kx}$ by the chain rule, and for $a^{x} = e^{(x\cdot{}\ln{} a)}$ it follows that $(a^x)' = \ln(a)a^x$ . This is why the exponential function describes all processes whose rate of change is proportional to the current amount.

Exam blueprint

Validity range

Holds for all real x; the simple form (eˣ)′ = eˣ requires base e. For other bases (aˣ)′ = ln(a)·aˣ, and for $e^{(kx)}$ the factor k comes from the chain rule.

Derivation steps

The difference quotient of eˣ contains eˣ as a factor.

1(e^(x+h) − eˣ)/h = eˣ·(e^h − 1)/h.
2As h → 0, $\frac{(e^{h} - 1)}{h}$ tends to 1, which is exactly what defines the base e; eˣ remains.

Rearrangements

Composite exponent

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

Inner derivative to the front, the exponential stays.

General base

\frac{d}{dx}\, a^{x} = \ln(a) \cdot a^{x}

Via $a^{x} = e^{(x\cdot{}\ln{} a)}$ ; for $a = e$ , $\ln{} a = 1$ .

Antiderivative

\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C

Integrating divides by k, differentiating multiplies by k.

Task variant

Differentiate $f(x) = 5e^{(-0.5x)}$ .

Inner derivative −0.5: f′ $(x) = 5\cdot{}(-0.5)\cdot{}e^{(-0.5x)} = -2.5e^{(-0.5x)}$ .
The function decreases everywhere, since $e^{(-0.5x)} > 0$ .

Show that $f(x) = e^{(3x)}$ satisfies f′ = 3f.

f′ $(x) = 3e^{(3x)} = 3\cdot{}f(x)$ .
In general $e^{(kx)}$ solves the differential equation f′ = k·f, the hallmark of exponential growth.

Common mistakes

Differentiating eˣ with the power rule: $x\cdot{}e^{(x-1)}$ .

The power rule applies to xⁿ (variable in the base), not to eˣ (variable in the exponent).

Forgetting the inner derivative 2 for $e^{(2x)}$ .

Chain rule: $(e^{(2x)})$ ′ = 2e^(2x).

Assuming (aˣ)′ = aˣ for any base.

Only for $a = e$ ; otherwise the factor ln a appears.

Exam context

Core of every exponential curve analysis: extrema, inflection points, tangents, growth models.

Formula cluster

Exponential functions

Derivative, antiderivative and growth models of the exponential function interlock.

(ln x)' = 1/x

Derivative of the Natural Logarithm

(f∘g)' = f'(g(x))·g'(x)

Chain Rule of Differentiation

∫xⁿ dx = xⁿ⁺¹/(n+1) + C

Antiderivatives and Basic Integrals

Worked example

$f(x) = 3e^{(2x)}$ : $f'(x) = 3\cdot2\cdot e^{2x} = 6e^{2x}$ (chain rule, inner derivative 2). At $x = 0$ : $f(0) = 3$ and $f'(0) = 6$ , the slope is k times the function value.

Applications

Exponential growth and decay, differential equations, population and capital models, curve analysis of exponential functions in final exams

How do you calculate with Derivative of the Exponential Function?

Here is how to work through a typical Derivative of the Exponential Function ((eˣ)' = eˣ) task step by step:

1
Task
Differentiate $f(x) = 5e^{(-0.5x)}$ .
Solution path
1. Inner derivative −0.5: f′ $(x) = 5\cdot{}(-0.5)\cdot{}e^{(-0.5x)} = -2.5e^{(-0.5x)}$ .
2. The function decreases everywhere, since $e^{(-0.5x)} > 0$ .
2
Task
Show that $f(x) = e^{(3x)}$ satisfies f′ = 3f.
Solution path
1. f′ $(x) = 3e^{(3x)} = 3\cdot{}f(x)$ .
2. In general $e^{(kx)}$ solves the differential equation f′ = k·f, the hallmark of exponential growth.

Frequently asked questions about Derivative of the Exponential Function

Why is the derivative of

e^{x}

again

e^{x}

The difference quotient of eˣ can be rewritten as eˣ·(e^h − 1)/h, because e^(x+h) = eˣ·e^h. The factor $\frac{(e^{h} - 1)}{h}$ does not depend on x; its limit as h → 0 is the slope of the exponential function at 0. The number e is defined precisely so that this limit equals 1: of all exponential functions aˣ, only the one with base $e \approx{} 2.71828$ has slope 1 at 0. That is why differentiating simply leaves eˣ. For any other base, (aˣ)′ = ln(a)·aˣ, the factor ln a measuring how much the base deviates from e. This self-reproduction makes the exponential function the natural model for growth whose speed is proportional to the current amount.

How do I differentiate

e^{(2x)}

or generally

e^{(kx)}

Use the chain rule: the outer function is the exponential, whose derivative is itself, the inner function is kx with derivative k. So $(e^{(kx)})$ ′ = k·e^(kx); the factor from the exponent moves to the front. Concretely: $(e^{(2x)})$ ′ = 2e^(2x), $(e^{(-0.5x)})$ ′ = −0.5·e^(−0.5x). With a general expression in the exponent the same principle applies: $(e^{(u(x))})$ ′ = u′ $(x)\cdot{}e^{(u(x))}$ , for example $(e^{(x^{2})})$ ′ = 2x·e^(x²). The classic exam mistake is forgetting the inner derivative and writing only $e^{(2x)}$ ; the result is then too small by the factor 2. Mnemonic: copy the exponential, multiply the inner derivative in front.

How do you differentiate

a^{x}

, for example

2^{x}

Rewrite the base: $a^{x} = e^{(x\cdot{}\ln{} a)}$ , since $e^{(\ln{} a)} = a$ . Now the chain rule applies with inner derivative ln a, so $(a^{x})$ ′ = ln(a)·a^x. For $2^{x}$ this means $(2^{x})$ ′ = ln(2)·2^x ≈ 0.693·2^x, for $10^{x}$ correspondingly $\ln{}(10)\cdot{}10^{x} \approx{} 2.303\cdot{}10^{x}$ . The factor ln a is the growth rate: for $a > e$ it exceeds 1, for $a < e$ it is smaller. The formula also shows why e is the "natural" base: only for $a = e$ does $\ln{} a = 1$ and the prefactor disappear. Exams like to use $2^{x}$ as a trap, because many reflexively write $2^{x}$ or $x\cdot{}2^{(x-1)}$ ; both are wrong.

How do I differentiate products like

x\cdot{}e^{x}

x^{2}\cdot{}e^{(-x)}

Here you combine product rule and chain rule. For f(x) = x·eˣ, f′(x) = 1·eˣ + x·eˣ = (1 + x)·eˣ. For $g(x) = x^{2}\cdot{}e^{(-x)}$ the product rule gives g′ $(x) = 2x\cdot{}e^{(-x)} + x^{2}\cdot{}(-1)\cdot{}e^{(-x)} = (2x - x^{2})\cdot{}e^{(-x)}$ ; the inner derivative −1 comes from the chain rule. Factoring out the exponential at the end is practical: since eˣ or $e^{(-x)}$ is never zero, you read the zeros of the derivative from the polynomial factor alone, here $2x - x^{2} = x(2 - x)$ with candidates $x = 0$ and $x = 2$ . Exactly this pattern, polynomial times exponential, is the most common function type in final-exam curve analysis.

Why does the exponential function describe growth and decay processes?+

Because its derivative is proportional to itself: $f(x) = c\cdot{}e^{(kx)}$ satisfies f′ = k·f. Exactly this characterizes natural processes where the rate of change depends on the current amount: the more bacteria there are, the more are added per hour; the more radioactive nuclei exist, the more decay per second. Positive k means growth, negative k decay. The exponential family is the only class of functions with this property, which is why it solves the differential equation f′ = k·f. In applications you determine k from data such as doubling time $(k =\frac{ \ln{} 2}{T_{D}})$ or half-life $(k = -\frac{\ln{} 2}{T_{H}})$ . Thus the differentiation rule becomes a direct modelling tool for biology, physics and finance.

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Which common mistake happens with Derivative of the Exponential Function?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Calculus Volume 1, §3.9: Derivatives of Exponential and Logarithmic Functions

Related formulas

\frac{d}{dx}(x^n) = n \cdot x^{n-1}

Power Rule of Differentiation

(\sin x)' = \cos x, \quad (\cos x)' = -\sin x

Derivatives of Sine and Cosine

More formulas: Calculus

e^{i\varphi} = \cos\varphi + i \cdot \sin\varphi

Euler's Formula

(f \cdot g)' = f' \cdot g + f \cdot g'

Product Rule of Differentiation

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n

Taylor Series

\left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^{2}}

Quotient Rule of Differentiation

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Derivative of the Exponential Function

The exponential function is its own derivative; for

e^{(kx)}

the chain rule adds the factor k.

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

e

Euler number

(\approx{} 2.71828)

dimensionless

x

Independent variable in the exponent

dimensionless

k

Constant factor in the exponent (growth rate)

dimensionless

Derivation & background

Exam blueprint

Validity range

Holds for all real x; the simple form (eˣ)′ = eˣ requires base e. For other bases (aˣ)′ = ln(a)·aˣ, and for $e^{(kx)}$ the factor k comes from the chain rule.

Derivation steps

The difference quotient of eˣ contains eˣ as a factor.

1(e^(x+h) − eˣ)/h = eˣ·(e^h − 1)/h.
2As h → 0, $\frac{(e^{h} - 1)}{h}$ tends to 1, which is exactly what defines the base e; eˣ remains.

How do you calculate with Derivative of the Exponential Function?

Here is how to work through a typical Derivative of the Exponential Function ((eˣ)' = eˣ) task step by step:

Task

Differentiate $f(x) = 5e^{(-0.5x)}$ .

Solution path

Inner derivative −0.5: f′ $(x) = 5\cdot{}(-0.5)\cdot{}e^{(-0.5x)} = -2.5e^{(-0.5x)}$ .
The function decreases everywhere, since $e^{(-0.5x)} > 0$ .

Task

Show that $f(x) = e^{(3x)}$ satisfies f′ = 3f.

Solution path

f′ $(x) = 3e^{(3x)} = 3\cdot{}f(x)$ .
In general $e^{(kx)}$ solves the differential equation f′ = k·f, the hallmark of exponential growth.

Frequently asked questions about Derivative of the Exponential Function

Why is the derivative of

e^{x}

again

e^{x}

How do I differentiate

e^{(2x)}

or generally

e^{(kx)}

How do you differentiate

a^{x}

, for example

2^{x}

How do I differentiate products like

x\cdot{}e^{x}

x^{2}\cdot{}e^{(-x)}

Why does the exponential function describe growth and decay processes?+