Derivative of e^x

The derivative of exe^x is exe^x itself—the unique exponential function used to model continuous growth, radioactive decay, and other rates of change.

ddxex=ex\frac{d}{dx}e^{x} = e^{x}

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What each symbol means

What Derivative of e^x takes
xx
ee
Derivative of e^x
SymbolMeaning
xxThe independent variable (usually representing time, distance, or another continuous quantity); if treated as a constant or coefficient instead of the exponent, you will lose the exponential relationship.
eeThe mathematical constant e2.718e \approx 2.718, the base of the natural exponential function and the unique base for which the derivative equals the original function; using a different base like 2 or 10 requires a different derivative formula.

When to use it

When you encounter an exponential function with base ee and need to find how fast it is changing at any point.

Level

Usually taught in: Calculus I

Worked examples

1. Derivative of a sum involving $e^x$

Problem

Find the derivative of f(x)=ex+3f(x) = e^x + 3.
  1. f(x)=ddx(ex+3)=ddxex+ddx3f'(x) = \frac{d}{dx}(e^x + 3) = \frac{d}{dx}e^x + \frac{d}{dx}3

    Use the sum rule to separate the derivative into two parts.

  2. ddxex=ex\frac{d}{dx}e^x = e^x

    Apply the exponential derivative rule: the derivative of exe^x is exe^x itself.

  3. ddx3=0\frac{d}{dx}3 = 0

    The derivative of any constant is zero.

  4. f(x)=ex+0=exf'(x) = e^x + 0 = e^x

    Combine both results to write the final answer.

Answer: f(x)=exf'(x) = e^x

Even though exe^x is added to a constant, the derivative removes the constant term entirely. This shows that when exe^x appears in a sum, it dominates the rate of change.

2. Derivative with a negative fractional exponent

Problem

Find the derivative of h(x)=ex/2h(x) = e^{-x/2}.
  1. h(x)=ddxex/2h'(x) = \frac{d}{dx}e^{-x/2}

    Write the derivative that we need to find.

  2. Let u=x2, then dudx=12\text{Let } u = -\frac{x}{2}, \text{ then } \frac{du}{dx} = -\frac{1}{2}

    Identify the inner function and compute its derivative, since we have a composition and will need the chain rule.

  3. ddxeu=eududx=ex/2(12)\frac{d}{dx}e^{u} = e^{u} \cdot \frac{du}{dx} = e^{-x/2} \cdot \left(-\frac{1}{2}\right)

    Apply the chain rule: multiply the derivative of eue^u (which is eue^u) by the derivative of the exponent 12-\frac{1}{2}.

  4. h(x)=12ex/2h'(x) = -\frac{1}{2}e^{-x/2}

    Simplify to get the final answer, keeping the negative sign.

Answer: h(x)=12ex/2h'(x) = -\frac{1}{2}e^{-x/2}

The chain rule is essential when the exponent is not simply xx; here the negative fraction requires careful attention to the sign and magnitude of the coefficient.

3. Exponential growth in a savings account

Problem

A savings account grows according to A(t)=2000e0.04tA(t) = 2000e^{0.04t} dollars, where tt is time in years. At what rate (in dollars per year) is the account growing after 10 years?
  1. A(t)=ddt(2000e0.04t)A'(t) = \frac{d}{dt}(2000e^{0.04t})

    Set up the derivative to find the rate of growth.

  2. A(t)=2000ddte0.04tA'(t) = 2000 \cdot \frac{d}{dt}e^{0.04t}

    Use the constant multiple rule to factor out 2000.

  3. Let u=0.04t, then dudt=0.04\text{Let } u = 0.04t, \text{ then } \frac{du}{dt} = 0.04

    Recognize that the exponent is a composition; set up the chain rule with the derivative of the inner function.

  4. ddte0.04t=e0.04t0.04=0.04e0.04t\frac{d}{dt}e^{0.04t} = e^{0.04t} \cdot 0.04 = 0.04e^{0.04t}

    Apply the chain rule: multiply the exponential derivative by the rate of change of the exponent.

  5. A(t)=20000.04e0.04t=80e0.04tA'(t) = 2000 \cdot 0.04e^{0.04t} = 80e^{0.04t}

    Multiply the constant by the coefficient from the exponent.

  6. A(10)=80e0.0410=80e0.4A'(10) = 80e^{0.04 \cdot 10} = 80e^{0.4}

    Substitute t=10t = 10 into the derivative function.

  7. A(10)801.492=119.4A'(10) \approx 80 \cdot 1.492 = 119.4

    Use a calculator or table: e0.41.492e^{0.4} \approx 1.492, so the rate is approximately $119.40\$119.40 per year.

Answer: A(10)=80e0.4119.40 dollars per yearA'(10) = 80e^{0.4} \approx 119.40 \text{ dollars per year}

Exponential models in real applications use base ee because it captures continuous compounding; the derivative tells us the instantaneous rate at which the account is growing at that exact moment.

Common mistakes

Where Derivative of e^x usually goes wrong
Answer came out wrong
Writing ddxe2x=e2x\frac{d}{dx}e^{2x} = e^{2x} and forgetting the chain rule coefficient.
Always multiply by the coefficient from the exponent: ddxe2x=e2x2=2e2x\frac{d}{dx}e^{2x} = e^{2x} \cdot 2 = 2e^{2x}.
Applying the power rule as if exe^x were a power function: ddxex=xex1\frac{d}{dx}e^x = x e^{x-1}.
Use the exponential derivative rule instead: ddxex=ex\frac{d}{dx}e^x = e^x.
Using ddx2x=2x\frac{d}{dx}2^x = 2^x without a coefficient, confusing other bases with base ee.
For other bases, use the formula ddxax=axln(a)\frac{d}{dx}a^x = a^x \ln(a); see Derivative of axa^x for details.
The mistakeWhy it is wrongThe fix
Writing ddxe2x=e2x\frac{d}{dx}e^{2x} = e^{2x} and forgetting the chain rule coefficient.The chain rule requires you to multiply by the derivative of the exponent; without it, the derivative is incomplete.Always multiply by the coefficient from the exponent: ddxe2x=e2x2=2e2x\frac{d}{dx}e^{2x} = e^{2x} \cdot 2 = 2e^{2x}.
Applying the power rule as if exe^x were a power function: ddxex=xex1\frac{d}{dx}e^x = x e^{x-1}.The power rule applies to xnx^n where xx is the base, but in exe^x, ee is the base and xx is the exponent—they have opposite roles, so different rules apply.Use the exponential derivative rule instead: ddxex=ex\frac{d}{dx}e^x = e^x.
Using ddx2x=2x\frac{d}{dx}2^x = 2^x without a coefficient, confusing other bases with base ee.The property that the derivative of exe^x equals exe^x is unique to the base ee and does not hold for other bases like 2, 10, or π\pi.For other bases, use the formula ddxax=axln(a)\frac{d}{dx}a^x = a^x \ln(a); see Derivative of axa^x for details.

Tips and when to use something else

  • The derivative of exe^x is the only exponential function whose derivative equals itself—this special property is why ee is called the natural base and is ubiquitous in calculus.
  • Whenever the exponent is not just xx alone (such as e3xe^{3x}, exe^{-x}, or ex2e^{x^2}), the chain rule is mandatory; the chain rule coefficient is never optional.
  • If your exponent is a more complex expression like ex2+3xe^{x^2 + 3x}, write out the chain rule carefully: multiply ex2+3xe^{x^2 + 3x} by ddx(x2+3x)=2x+3\frac{d}{dx}(x^2 + 3x) = 2x + 3.
  • When exe^x appears in a product (such as xexxe^x or exsin(x)e^x \sin(x)), switch to the product rule instead—see Product Rule for how to handle these cases.

Frequently asked questions

Why is the derivative of exe^x equal to exe^x itself and not something else?
This is a defining property of ee. The number e2.718e \approx 2.718 is the unique base for which the exponential function and its derivative are identical. This arises from the deeper definition: ee is the limit of (1+1/n)n(1 + 1/n)^n as nn \to \infty, and this same limit governs how exponential growth behaves. Bases like 2 or 10 do not have this property.
Do I have to use the chain rule for exe^x by itself, or only when the exponent is more complex?
For exe^x alone, you do not need the chain rule; the derivative is simply exe^x. You only invoke the chain rule when the exponent is something other than just xx (such as e2xe^{2x}, exe^{-x}, or ex/3e^{x/3}).
Can I use this rule if my exponent is a fraction or a decimal, like e0.05te^{0.05t}?
Yes. Treat the coefficient as you would any other constant in the exponent and apply the chain rule. For example, ddte0.05t=e0.05t0.05=0.05e0.05t\frac{d}{dt}e^{0.05t} = e^{0.05t} \cdot 0.05 = 0.05e^{0.05t}. Fractions and decimals follow the same rule.
What should I do if I see the derivative of a base like 2 or 10 instead of ee?
Use the formula for the derivative of axa^x, which is ddxax=axln(a)\frac{d}{dx}a^x = a^x \ln(a). For example, ddx10x=10xln(10)\frac{d}{dx}10^x = 10^x \ln(10). The natural logarithm ln\ln appears because ee is still fundamental; it emerges whenever we take derivatives of other bases.

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Reviewed 2026-09-18