practicecalculus
Exponential and log derivatives

Derivative of e^(−x): Formula, Steps and Examples

The derivative of e^(−x) is −e^(−x). See the chain rule steps, a check with the quotient rule, higher derivatives, a graph and examples like xe^(−x).

Exponential and log derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,e^{-x} = -e^{-x}
Derivative
-e^{-x}
Second derivative
e^{-x}
nth derivative
(-1)^n e^{-x}
Same function as
\dfrac{1}{e^x}
Integral version
\int e^{-x}\,dx = -e^{-x} + C
On this page
  1. Step by step with the chain rule
  2. Check with the quotient rule
  3. Graph of e^(−x) and its derivative
  4. Higher derivatives
  5. Worked examples
  6. Common mistakes

The derivative of e^{-x} is -e^{-x}. The exponential stays the same, and the derivative of the exponent -x, which is -1, puts a minus sign in front.

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

Step by step with the chain rule

Think of e^{-x} as e^u with u = -x. The chain rule gives

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

The outer step uses the fact that the derivative of eˣ is e^x.

Check with the quotient rule

Since e^{-x} = \frac{1}{e^x}, the quotient rule should agree:

\frac{d}{dx}\frac{1}{e^x} = \frac{0\cdot e^x - 1\cdot e^x}{(e^x)^2} = \frac{-e^x}{e^{2x}} = -e^{-x}

Same answer, with more work. The chain rule is the quick way.

Graph of e^(−x) and its derivative

Graph of y = e^(−x) and its derivative y = −e^(−x)
The function e^{-x} is always positive and always falling. Its derivative is the mirror image below the x-axis.

The graph of e^{-x} is the graph of e^x reflected across the y-axis. Reflecting left to right turns every rising slope into a falling one, so the derivative is negative. As x \to \infty both the function and its slope shrink toward 0. This is the shape of exponential decay.

Higher derivatives

Each derivative brings out another -1, so the signs alternate:

Derivative Result
f'(x) -e^{-x}
f''(x) e^{-x}
f'''(x) -e^{-x}
f^{(n)}(x) (-1)^n e^{-x}

Since f''(x) = f(x), the function e^{-x} solves the differential equation y'' = y, just like e^x does. And f'(x) = -f(x) says it shrinks at a rate equal to its own size, which is the defining rule of exponential decay.

Worked examples

Example 1: product rule. For f(x) = xe^{-x}, the product rule gives

f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)

So f rises for x < 1, falls for x > 1, and has a maximum of f(1) = \frac{1}{e} \approx 0.368.

Example 2: exponential decay. A quantity A(t) = 50e^{-0.2t} changes at the rate

A'(t) = 50\cdot(-0.2)e^{-0.2t} = -10e^{-0.2t}

At t = 0 it is falling by 10 units per unit time.

Example 3: a sum. For g(x) = e^x + e^{-x},

g'(x) = e^x - e^{-x}

Divide both by 2 and you get the derivative of cosh x, which is \sinh x.

Common mistakes

  1. Dropping the minus sign. \frac{d}{dx}e^{-x} = e^{-x} is wrong. The inside derivative is -1.
  2. Writing -xe^{-x - 1}. That uses the power rule on an exponential.
  3. Losing track of signs in higher derivatives. Even derivatives are positive, odd ones are negative.
  4. Getting the integral sign wrong. \int e^{-x}\,dx = -e^{-x} + C, not e^{-x} + C. See the integral of e^(−x).

For more chain rule practice with exponentials, try derivatives of exponential and logarithmic functions.

Practice it now

Chain rule: e and ln

Solve real problems on this topic with instant feedback, hints and a step-by-step walkthrough for every answer. Free, no sign-up needed to start.

Practice this lesson

Frequently asked questions

What is the derivative of e^(−x)?

The derivative of e^{-x} is -e^{-x}.

What is the derivative of 1/e^x?

Since \frac{1}{e^x} = e^{-x}, the derivative is -e^{-x} = -\frac{1}{e^x}.

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

Differentiate -e^{-x} again: -(-e^{-x}) = e^{-x}. The second derivative equals the original function.

What is the derivative of x·e^(−x)?

By the product rule, e^{-x} - xe^{-x} = e^{-x}(1 - x).

Written by the PracticeCalculus team

We build interactive calculus lessons and write every page to be clear and correct. Spotted something off? Tell us and we will fix it.