Derivatives of Exponential Functions

The derivative of an exponential function depends on its base. The derivative of $e^x$ is $e^x$ itself. For any other base $a \gt 0$ , the derivative of $a^x$ is $a^x \ln a$ . With an inside function u, use the chain rule: $\frac{d}{dx}e^u = e^u \cdot u'$ and $\frac{d}{dx}a^u = a^u \ln a \cdot u'$ .

Before this lesson: Chain Rule, The Number e and Logarithms

Why e^x is its own derivative

Every exponential curve $y = a^x$ has a slope that is proportional to its height: the higher the curve, the steeper it is. For $y = 2^x$ , the slope is about 0.69 times the height. For $y = 3^x$ , it is about 1.10 times the height.

The number $e \approx 2.718$ is the base where that factor is exactly 1. So at every point on $y = e^x$ , the slope equals the height. At (0, 1) the slope is 1, and at (1, e) the slope is e.

Graph of y = e^x for x from −3 to 2. Dashed tangent lines are drawn at (0, 1), where the slope is 1, and at (1, e), where the slope is e, about 2.72. At each point the slope equals the height of the curve. — On y = eˣ, the slope at every point equals the y-value there.

The factor for base a is the natural log, ln a. That is where the formula $a^x \ln a$ comes from, and since $\ln e = 1$ , it agrees with the rule for $e^x$ .

Exponential derivative rules

Function	Derivative	With an inside function u
$e^x$	$e^x$	$e^u \cdot u'$
$a^x$	$a^x \ln a$	$a^u \ln a \cdot u'$
$k e^x$ (k a constant)	$k e^x$	$k e^u \cdot u'$

Do not use the power rule here. The power rule is for $x^n$ , where the variable is in the base. In $e^x$ and $a^x$ , the variable is in the exponent.

Derivatives of exponential functions examples

Example 1: A sum with e^x

Differentiate $y = 4e^x - x^2$ .

y' = 4e^x - 2x

The derivative of eˣ is eˣ, so 4eˣ stays 4eˣ. Power rule on x².

Answer: $y' = 4e^x - 2x$ .

Example 2: Chain rule with e

Differentiate $y = e^{3x^2}$ .

u = 3x^2, \; u' = 6x

The inside function is the exponent, 3x².

y' = e^{3x^2} \cdot 6x

The derivative of eᵘ is eᵘ · u′ (see chain rule). Keep the exponent the same.

y' = 6x\,e^{3x^2}

Write the factor in front.

Answer: $y' = 6x\,e^{3x^2}$ .

Example 3: Product rule with e^x

Differentiate $y = x^2 e^x$ and factor the answer.

y' = (2x)(e^x) + (x^2)(e^x)

Product rule: the derivative of x² is 2x and of eˣ is eˣ (see product rule).

y' = e^x(x^2 + 2x)

Factor out eˣ.

y' = x e^x(x + 2)

Factor out x (see factoring out the GCF).

Answer: $y' = xe^x(x + 2)$ . Since $e^x$ is never 0, the factored form shows that y′ = 0 only at x = 0 and x = −2.

Example 4: A base other than e

Differentiate $y = 5^x$ and find the slope at x = 2.

y' = 5^x \ln 5

Use the rule for aˣ with a = 5.

y'(2) = 5^2 \ln 5 = 25\ln 5

Substitute x = 2.

25\ln 5 \approx 25(1.6094) \approx 40.24

Use ln 5 ≈ 1.6094.

Answer: $y' = 5^x \ln 5$ , and the slope at x = 2 is $25\ln 5 \approx 40.24$ .

Example 5: Word problem with bacteria

A bacteria colony has $P(t) = 500e^{0.4t}$ cells after t hours. How fast is it growing at t = 5 hours?

P'(t) = 500e^{0.4t} \cdot 0.4

Chain rule: the exponent 0.4t has derivative 0.4.

P'(t) = 200e^{0.4t}

Multiply 500 · 0.4.

P'(5) = 200e^{2}

Substitute t = 5: 0.4 · 5 = 2.

200e^2 \approx 200(7.389) \approx 1478

Use e² ≈ 7.389.

Answer: at 5 hours, the colony grows by about 1,478 cells per hour. Notice that $P'(t) = 0.4P(t)$ : the growth rate is always 40% of the current size per hour, which is what exponential growth means (see exponential growth and decay).

Common mistakes

Mistake	Example of the error	Fix
Using the power rule on e^x	$\frac{d}{dx}e^x = xe^{x-1}$	The variable is in the exponent: $\frac{d}{dx}e^x = e^x$ .
Forgetting the chain rule	$\frac{d}{dx}e^{3x^2} = e^{3x^2}$	Multiply by the exponent’s derivative, 6x.
Changing the exponent	$\frac{d}{dx}e^{3x^2} = e^{6x}$	Keep the exponent the same: $6x\,e^{3x^2}$ .
Forgetting ln a for other bases	$\frac{d}{dx}5^x = 5^x$	Only base e works that way: $5^x \ln 5$ .
Treating e as a variable	$\frac{d}{dx}e^2 = 2e$	$e^2$ is a constant, so its derivative is 0.

Related methods

Rewriting a^x with base e

Any exponential can be written with base e: $a^x = e^{x\ln a}$ . Then the chain rule gives $e^{x\ln a} \cdot \ln a = a^x \ln a$ , which is the rule above.

Logarithmic derivatives

The derivative of $\ln x$ is the next lesson, Derivatives of Logarithmic Functions. It comes from this one: since $\ln x$ is the inverse of $e^x$ , the rule for derivatives of inverse functions gives $\frac{1}{x}$ .

Practice problems

Differentiate $y = 7e^x$ .
Differentiate $y = e^{5x}$ .
Differentiate $y = e^{-x^2}$ .
Differentiate $y = 2^x$ .
Differentiate $y = xe^{2x}$ and factor the answer.
Differentiate $y = \frac{e^x}{x}$ .
Find the tangent line to $y = e^x$ at x = 0.
Differentiate $y = 3^{x^2}$ .
Find where $y = xe^{-x}$ has a horizontal tangent line.
A car worth $24,000 loses value so that it is worth $V(t) = 24000(0.85)^t$ dollars after t years. How fast is its value changing at t = 2?

Show answers

y' = 7e^x

A constant times eˣ keeps the constant.

Answer: $y' = 7e^x$

y' = e^{5x} \cdot 5

Chain rule: the exponent has derivative 5.

Answer: $y' = 5e^{5x}$

y' = e^{-x^2} \cdot (-2x)

Chain rule: the exponent −x² has derivative −2x.

Answer: $y' = -2x\,e^{-x^2}$

y' = 2^x \ln 2

Rule for aˣ with a = 2.

Answer: $y' = 2^x \ln 2$

y' = 1 \cdot e^{2x} + x \cdot 2e^{2x}

Product rule, with the chain rule on e²ˣ.

y' = e^{2x}(1 + 2x)

Factor out e²ˣ.

Answer: $y' = e^{2x}(1 + 2x)$

y' = \frac{e^x \cdot x - e^x \cdot 1}{x^2}

Quotient rule (see quotient rule).

y' = \frac{e^x(x - 1)}{x^2}

Factor out eˣ in the numerator.

Answer: $y' = \frac{e^x(x - 1)}{x^2}$

y(0) = e^0 = 1

Find the point (0, 1).

y'(0) = e^0 = 1

The slope is also e⁰ = 1.

y - 1 = 1(x - 0)

Point-slope form (see point-slope form).

y = x + 1

Add 1 to both sides.

Answer: $y = x + 1$

y' = 3^{x^2} \ln 3 \cdot 2x

Rule for aᵘ with u = x² and u′ = 2x.

Answer: $y' = 2x \cdot 3^{x^2}\ln 3$

y' = 1 \cdot e^{-x} + x \cdot (-e^{-x})

Product rule; the derivative of e⁻ˣ is −e⁻ˣ.

y' = e^{-x}(1 - x)

Factor out e⁻ˣ.

1 - x = 0 \Rightarrow x = 1

e⁻ˣ is never 0, so set the other factor equal to 0.

Answer: x = 1, at the point (1, 1/e)

10.

V'(t) = 24000(0.85)^t \ln 0.85

Rule for aˣ with a = 0.85.

V'(2) = 24000(0.7225)\ln 0.85

Substitute t = 2: 0.85² = 0.7225.

V'(2) = 17340 \ln 0.85 \approx 17340(-0.16252)

Multiply, and use ln 0.85 ≈ −0.16252.

V'(2) \approx -2818

Multiply.

Answer: after 2 years, the car is losing value at about $2,818 per year.

Derivatives of exponential functions FAQ

What is the derivative of e^x?

The derivative of $e^x$ is $e^x$ . It is the only basic function (apart from constant multiples of it) that is its own derivative.

What is the derivative of a^x?

$\frac{d}{dx}a^x = a^x \ln a$ . For example, $\frac{d}{dx}2^x = 2^x \ln 2 \approx 0.693 \cdot 2^x$ .

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

By the chain rule, $\frac{d}{dx}e^{2x} = 2e^{2x}$ . Multiply by the derivative of the exponent.

Why can’t you use the power rule on e^x?

The power rule is for a variable base raised to a constant power, like $x^3$ . In $e^x$ , the base is a constant and the variable is in the exponent, so a different rule applies.

Summary

$\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx}a^x = a^x \ln a$ .
With an inside function: $\frac{d}{dx}e^u = e^u u'$ .
Keep the exponent the same, then multiply by its derivative.
Never use the power rule when the variable is in the exponent.
For growth models, $P' = kP$ : the rate is proportional to the amount.

Previous lesson: Derivatives of Trig Functions

Next lesson: Derivatives of Logarithmic Functions

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