Derivative of e^ax
2026-02-28 23:53 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-e%5Eax

We can derive the derivative of e^ax using proofs. To show this, we will use the exponential properties along with the rules of differentiation.

There are several methods we use to prove this, such as:

By First Principle Using Chain Rule We will now demonstrate that the differentiation of e^ax results in ae^ax using the above-mentioned methods:

By First Principle The derivative of e^ax can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of e^ax using the first principle, we will consider f(x) = e^ax. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

Given that f(x) = e^ax, we write f(x + h) = e^(a(x + h)).

Substituting these into equation (1), f'(x) = limₕ→₀ [e^(a(x + h)) - e^ax] / h = limₕ→₀ [e^ax (e^(ah) - 1)] / h = e^ax limₕ→₀ [e^(ah) - 1] / h

We now make a substitution: let u = ah, then as h approaches 0, u approaches 0. f'(x) = e^ax limᵤ→₀ [e^u - 1] / (u/a) = ae^ax limᵤ→₀ [e^u - 1] / u

Using limit formulas, limᵤ→₀ [e^u - 1] / u = 1. f'(x) = ae^ax. Hence, proved.

Using Chain Rule To prove the differentiation of e^ax using the chain rule, We use the formula: Let u = ax Then e^ax = e^u

By chain rule: d/dx [e^u] = e^u · du/dx = e^ax · a Therefore, d/dx (e^ax) = ae^ax.