Derivative of e^(2x)
e^(2x) is the simplest possible chain-rule exponential — and a useful introduction to differentiating composite exponentials.
Step-by-step solution
The exponent isn't plain x; it's the inner function u = 2x. That means chain rule: differentiate the outside, then multiply by the derivative of the inside.
Outside: d/du[eu] = eu. Inside: d/dx[2x] = 2.
f′(x) = e2x · 2 = 2e2x.
Why it works
The pattern generalizes instantly: for any constant k, d/dx[ekx] = k·ekx. The exponential survives unchanged and the inner slope k pops out front. This one-liner is why e^(kx) solves the differential equation f′ = k·f, which models everything from bacterial growth (k > 0) to drug elimination in the bloodstream (k < 0).
It also explains half-life and doubling-time intuition: e^(2x) is e^x running at double speed, so at every point its slope is twice as steep relative to the same height. Compressing an exponential horizontally is indistinguishable from scaling its growth rate.
Domain and repeated differentiation
The function exp(2x) is defined and differentiable for all real x. Its derivative is positive everywhere, so it is strictly increasing. At x = 0 the value is 1 and the slope is 2, giving the tangent line y = 1 + 2x.
Each further differentiation multiplies by another 2: f″(x) = 4 exp(2x), and the nth derivative is 2ⁿ exp(2x) for a nonnegative integer n. For exp(g(x)), the first derivative is g′(x) exp(g(x)) wherever g is differentiable.
Common mistakes
- Answering e^(2x) with no factor of 2 — forgetting the chain rule entirely.
- Writing 2x·e^(2x−1), a hybrid of the power rule and the exponential rule that is valid for neither.
- Putting the 2 in the exponent of the answer, e.g. e^(4x). The inner derivative multiplies out front; it doesn't re-enter the exponent.
Reading e^(2x) correctly
Here e is the constant natural-logarithm base, and the entire 2x is the exponent. The expressions exp(2x) and e^(2x) mean the same thing. They differ from 2exp(x), whose derivative is 2exp(x), and from exp(x²), whose derivative is 2x exp(x²).
For exp(ax+b), with fixed real constants a and b, the derivative is a exp(ax+b). If a=0, the function is constant and the derivative is zero. A fixed shift b changes the function value but adds no extra multiplier.
Practice problems
Differentiate e^(−3x)
Answer: −3e^(−3x) — the constant k can be negative.
Differentiate 4e^(2x)
Answer: 8e^(2x).
Differentiate e^(x²)
Answer: 2x·e^(x²) — the inner derivative is now a function, not a constant.