Derivative of e^(2x): 2e^(2x) Step by Step

Quick answer:

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

The exponential keeps its shape, and the 2 from the exponent comes out in front. This guide shows three different ways to see why, generalizes the result to \( e^{kx} \), and works through examples that combine it with the product, quotient and chain rules, including a full exam-style max/min problem.

Why it works: the intuition

The graph of \( e^{2x} \) is the graph of \( e^x \) squeezed horizontally by a factor of 2. Whatever \( e^x \) does over one unit of \( x \), \( e^{2x} \) does over half a unit. Squeezing a graph sideways by 2 makes every slope twice as steep, and the slope of \( e^x \) at any height equals that height. So the slope of \( e^{2x} \) is twice its height: \( 2e^{2x} \).

There is also a growth picture. A quantity that obeys \( y = e^{2x} \) grows at a rate proportional to its current size, with constant of proportionality 2. Doubling the constant in the exponent doubles the relative growth rate. That is the whole content of the formula.

Why: the chain rule in one line

The exponential function is its own derivative, \( \frac{d}{dx}e^{u} = e^{u} \). When the exponent is a function \( u = g(x) \), the chain rule multiplies by \( g'(x) \):

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

With \( g(x) = 2x \), \( g'(x) = 2 \), so the derivative is \( 2e^{2x} \). In words: copy the exponential exactly as it is, then multiply by the derivative of the exponent.

Check without the chain rule

Since \( e^{2x} = e^{x}\cdot e^{x} \), the product rule gives \( e^{x}e^{x} + e^{x}e^{x} = 2e^{2x} \). Same answer.

A third route uses the rule for other bases. Since \( e^{2x} = (e^2)^x \), it is an exponential with base \( e^2 \). The derivative of \( a^x \) is \( a^x\ln a \), and \( \ln(e^2) = 2 \), so once again the derivative is \( 2e^{2x} \). Three methods, one answer, which is a good sign you understand the rule rather than just memorizing it.

The general pattern: e^(kx)

$$\frac{d}{dx}e^{kx} = k\,e^{kx}$$

• \( \frac{d}{dx}e^{5x} = 5e^{5x} \)
• \( \frac{d}{dx}e^{-x} = -e^{-x} \)
• \( \frac{d}{dx}e^{x/2} = \frac12 e^{x/2} \)

This is why \( e^{kx} \) models exponential growth and decay: the rate of change is proportional to the amount present. A positive \( k \) gives growth, a negative \( k \) gives decay, and the size of \( k \) sets how fast.

More worked examples

1. \( e^{x^2} \). The exponent’s derivative is \( 2x \):

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

Be careful to keep \( x^2 \) in the exponent unchanged; only the factor out front is new.

2. \( x e^{2x} \). Two factors are multiplied, so use the product rule. The derivative of \( x \) is 1 and the derivative of \( e^{2x} \) is \( 2e^{2x} \):

$$\frac{d}{dx}\big(xe^{2x}\big) = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x)$$

3. \( e^{2x}\sin x \). Product rule again, with the chain rule inside the first term:

$$\frac{d}{dx}\big(e^{2x}\sin x\big) = 2e^{2x}\sin x + e^{2x}\cos x$$

4. \( e^{\sin x} \). The exponent is \( \sin x \), whose derivative is \( \cos x \), so the answer is \( \cos x\cdot e^{\sin x} \).

5. A quotient: \( \frac{e^{2x}}{x} \). Use the quotient rule and factor out the exponential at the end:

$$\frac{d}{dx}\frac{e^{2x}}{x} = \frac{2xe^{2x} - e^{2x}}{x^2} = \frac{e^{2x}(2x - 1)}{x^2}$$

6. Two layers: \( (e^{2x} + 1)^3 \). The outside is a cube and the inside is \( e^{2x} + 1 \), whose derivative is \( 2e^{2x} \):

$$\frac{d}{dx}(e^{2x} + 1)^3 = 6e^{2x}(e^{2x} + 1)^2$$

7 (exam level): find the minimum of \( xe^{2x} \). From Example 2 the derivative is \( e^{2x}(1 + 2x) \). The exponential is never zero, so the derivative is zero only when \( 1 + 2x = 0 \), at \( x = -\frac12 \). The derivative is negative to the left and positive to the right, so this is a minimum, with value

$$-\frac12 e^{-1} = -\frac{1}{2e}$$

This “factor out the exponential, then solve the rest” move is standard when you look for critical points of exponential functions.

Tangent line at x = 0

At \( x = 0 \), \( e^{2x} = 1 \) and the slope is \( 2e^0 = 2 \). So the tangent line is \( y = 1 + 2x \). For small \( x \), this gives the quick estimate \( e^{2x} \approx 1 + 2x \); for example, \( e^{0.02} \approx 1.02 \). The true value is about 1.0202, so the tangent line is off by only about 0.0002. The estimate is always a little low, because \( e^{2x} \) is concave up and its graph bends above every tangent line. The further you move from 0, the bigger that gap becomes, so the shortcut is best for exponents close to zero.

Higher derivatives

Each differentiation multiplies by another 2:

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

So the second derivative is \( 4e^{2x} \) and the tenth is \( 1024e^{2x} \). The second derivative is always positive, so \( e^{2x} \) is concave up everywhere.

Growth, doubling time and differential equations

The function \( y = Ce^{2x} \) is the general solution of the differential equation \( y' = 2y \): the derivative formula says exactly that the rate equals twice the amount. If \( x \) is time, the amount doubles whenever \( 2x \) increases by \( \ln 2 \), so the doubling time is

$$\frac{\ln 2}{2} \approx 0.347$$

time units. You can explore other rates with the exponential growth calculator.

Where it’s used

Functions of the form \( e^{kx} \) are the standard model for anything that changes at a rate proportional to its size, so the derivative on this page appears across science:

• Populations and finance. Continuous growth, such as money earning continuously compounded interest or a bacteria culture with plenty of food, follows \( Ce^{kt} \). The derivative gives the growth rate at any moment.
• Radioactive decay and cooling. A negative constant, as in \( e^{-kt} \), gives decay. The derivative is negative and shrinks in size as the quantity runs out.
• Differential equations. Guessing a solution of the form \( e^{rx} \) is the first step in solving many linear differential equations, because differentiating only multiplies by \( r \). That turns calculus into algebra.
• Hyperbolic functions. Practice problem 5 below is the hyperbolic sine of \( 2x \) in disguise, and its derivative uses exactly the rule from this page.
• Series. Because every derivative of \( e^{2x} \) at 0 equals \( 2^n \), its Taylor series is easy to write down: the coefficients are \( \frac{2^n}{n!} \).

How to check your answer

Two quick checks catch most errors. First, evaluate at \( x = 0 \): the slope of \( e^{kx} \) there must be \( k \), so if your formula gives something else, look for a missing chain rule factor. Second, look at the sign. A decaying exponential like \( e^{-3x} \) must have a negative derivative, and a growing one a positive derivative. If the signs disagree, recheck the derivative of the exponent.

Common mistakes

• Using the power rule. Writing \( \frac{d}{dx}e^{2x} = 2x\,e^{2x-1} \) applies the power rule, which only works when the base is the variable and the exponent is a constant. Here the variable is in the exponent.
• Forgetting the 2. Answering \( e^{2x} \) treats the exponent as plain \( x \). The chain rule factor is required.
• Putting \( x \) in the factor. \( 2xe^{2x} \) is wrong; the derivative of \( 2x \) is the constant 2, not \( 2x \).
• Confusing \( e^{2x} \) with \( e^{x^2} \). \( (e^x)^2 = e^{2x} \), not \( e^{x^2} \). Their derivatives are \( 2e^{2x} \) and \( 2xe^{x^2} \).
• Multiplying when integrating. For the integral you divide by 2 instead: \( \int e^{2x}\,dx = \frac12 e^{2x} + C \).

Try it yourself

Type any exponential expression below to watch the chain and product rules at work, or open the full derivative calculator.

Practice problems

Try these before checking the answers.

1. \( \frac{d}{dx}e^{-3x} \)
2. \( \frac{d}{dx}e^{x^3} \)
3. \( \frac{d}{dx}\,x^2e^{2x} \)
4. \( \frac{d}{dx}e^{2x + 3} \)
5. \( \frac{d}{dx}\frac{e^{2x} - e^{-2x}}{2} \)

Answers: (1) \( -3e^{-3x} \); (2) \( 3x^2e^{x^3} \); (3) \( 2xe^{2x} + 2x^2e^{2x} \); (4) \( 2e^{2x+3} \); (5) \( e^{2x} + e^{-2x} \).

FAQ

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

\( \int e^{2x}\,dx = \frac12 e^{2x} + C \): divide by the coefficient instead of multiplying. It is a one-step u-substitution with \( u = 2x \). Over \( [0, 1] \) it gives \( \frac{e^2 - 1}{2} \).

What is the derivative of 2^x?

\( 2^x\ln 2 \). For a base other than \( e \), multiply by the natural log of the base.

Is the derivative of e^(2x) equal to e^(2x)?

No. Only \( e^x \) is exactly its own derivative. For \( e^{2x} \) the chain rule adds a factor of 2.

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

Differentiate \( 2e^{2x} \) once more to get \( 4e^{2x} \).

Related: derivative of x^x, derivative of ln x.

Further reading

• OpenStax Calculus Volume 1, 3.9 Derivatives of Exponential and Logarithmic Functions — the textbook section on exponential derivatives with exercises and answers.
• Paul’s Online Notes: Derivatives of Exponential and Logarithm Functions — why \( e^x \) is its own derivative and how other bases work.
• Paul’s Online Notes: Chain Rule — many more composite examples, including exponentials of functions.