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Exponential and log derivatives

Derivative of e^(2x): Formula, Steps and Examples

The derivative of e^(2x) is 2e^(2x). See the chain rule steps, a second proof using (eˣ)², higher derivatives, a graph and worked examples.

Exponential and log derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,e^{2x} = 2e^{2x}
Derivative
2e^{2x}
Second derivative
4e^{2x}
nth derivative
2^n e^{2x}
General form
\dfrac{d}{dx}e^{kx} = ke^{kx}
Integral version
\int e^{2x}\,dx = \frac{1}{2}e^{2x} + C
On this page
  1. Step by step with the chain rule
  2. A second proof without the chain rule
  3. Graph of e^(2x) and its derivative
  4. Higher derivatives
  5. Worked examples
  6. Common mistakes

The derivative of e^{2x} is 2e^{2x}. The exponential stays exactly the same, and the derivative of the exponent, 2, comes out in front.

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

Step by step with the chain rule

Write e^{2x} as an outer function e^u with inside u = 2x. The chain rule says

\frac{d}{dx}e^u = e^u\cdot u'
  1. The outer derivative: the derivative of eˣ is itself, so e^u stays as e^{2x}.
  2. The inside derivative: \frac{d}{dx}(2x) = 2.
  3. Multiply: e^{2x}\cdot 2 = 2e^{2x}.

The same steps give the general rule \frac{d}{dx}e^{kx} = ke^{kx}. For example, \frac{d}{dx}e^{5x} = 5e^{5x} and \frac{d}{dx}e^{x/3} = \frac{1}{3}e^{x/3}.

A second proof without the chain rule

Since e^{2x} = (e^x)^2 = e^x\cdot e^x, the product rule works too:

\frac{d}{dx}(e^x\cdot e^x) = e^x\cdot e^x + e^x\cdot e^x = 2e^{2x}

Both methods agree, which is a nice check.

Graph of e^(2x) and its derivative

Graph of y = e^(2x) and its derivative y = 2e^(2x)
The derivative is always exactly twice the height of the function. At x = 0 the function is 1 and the slope is 2.

The tangent line at x = 0 has slope 2 and passes through (0, 1), so it is y = 2x + 1.

Higher derivatives

Each time you differentiate, another factor of 2 comes out:

Derivative Result
f'(x) 2e^{2x}
f''(x) 4e^{2x}
f'''(x) 8e^{2x}
f^{(n)}(x) 2^n e^{2x}

This pattern is why y = e^{2x} solves the differential equation y'' = 4y. It also explains the first derivative on its own: y' = 2y says the function grows at a rate equal to twice its current size, which is the rule behind exponential growth with rate constant 2.

Worked examples

Example 1: product rule. For f(x) = xe^{2x},

f'(x) = 1\cdot e^{2x} + x\cdot 2e^{2x} = e^{2x}(1 + 2x)

Since e^{2x} > 0, the only critical point is where 1 + 2x = 0, at x = -\frac{1}{2}.

Example 2: a shifted exponent. For g(x) = e^{2x + 1}, the inside derivative is still 2:

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

Example 3: a constant multiple. For h(x) = 5e^{2x} - e^{-2x},

h'(x) = 10e^{2x} + 2e^{-2x}

The second term flips sign because \frac{d}{dx}(-2x) = -2.

Common mistakes

  1. Forgetting the 2. \frac{d}{dx}e^{2x} = e^{2x} is the most common slip. Always multiply by the inside derivative.
  2. Bringing the exponent down like a power. 2x\,e^{2x - 1} is wrong; the power rule does not apply.
  3. Writing 2e^{x}. The exponent does not change. It stays 2x.
  4. Mixing up the derivative and the integral. The derivative multiplies by 2; the integral divides by 2.

For the reverse direction, see the integral of e^(2x). For more practice, try derivatives of exponential and logarithmic functions.

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Frequently asked questions

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

The derivative of e^{2x} is 2e^{2x}, by the chain rule.

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

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

What is the derivative of x·e^(2x)?

By the product rule, e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x).

Is the derivative of e^(2x) equal to 2x·e^(2x − 1)?

No. That treats e^{2x} like a power of x. The exponent contains the variable, so use the exponential rule with the chain rule.

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