Last updated on July 16th, 2025
We use the derivative of e^ax, which is ae^ax, as a measuring tool for how the exponential function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of e^ax in detail.
We now understand the derivative of e^ax. It is commonly represented as d/dx (e^ax) or (e^ax)', and its value is ae^ax. The function e^ax has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: (e^ax is an exponential function where e is the base of the natural logarithm).
Chain Rule: Rule for differentiating e^ax (since it involves a composite function).
Constant Multiple Rule: When differentiating constant multiples of functions.
The derivative of e^ax can be denoted as d/dx (e^ax) or (e^ax)'.
The formula we use to differentiate e^ax is: d/dx (e^ax) = ae^ax (or) (e^ax)' = ae^ax
The formula applies to all x.
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.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like e^ax.
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.
For the nth derivative of e^ax, we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).
When x = 0, the derivative of e^ax = ae^(a*0), which is a. The derivative of e^0 is 0 because e^0 equals 1, and the derivative of a constant is 0.
Students frequently make mistakes when differentiating e^ax. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (e^ax · ax).
Here, we have f(x) = e^ax · ax. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^ax and v = ax.
Let’s differentiate each term, u′ = d/dx (e^ax) = ae^ax v′ = d/dx (ax) = a
Substituting into the given equation, f'(x) = (ae^ax)(ax) + (e^ax)(a) = ae^ax(ax) + ae^ax
Let’s simplify terms to get the final answer, f'(x) = a^2xe^ax + ae^ax
Thus, the derivative of the specified function is a^2xe^ax + ae^ax.
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
The growth of a bacteria population is represented by the function y = e^(0.1x), where y represents the population at time x. If x = 5 hours, measure the growth rate of the population.
We have y = e^(0.1x) (growth of the bacteria population)...(1)
Now, we will differentiate the equation (1). Take the derivative e^(0.1x): dy/dx = 0.1e^(0.1x)
Given x = 5 (substitute this into the derivative), dy/dx = 0.1e^(0.5)
Hence, we get the growth rate of the population at time x = 5 as 0.1e^(0.5).
We find the growth rate of the population at x = 5, which means that at a given point, the population's size is increasing at a rate proportional to e^(0.5).
Derive the second derivative of the function y = e^(2x).
The first step is to find the first derivative, dy/dx = 2e^(2x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2e^(2x)] = 2 * 2e^(2x) = 4e^(2x)
Therefore, the second derivative of the function y = e^(2x) is 4e^(2x).
We use the step-by-step process, where we start with the first derivative. We then differentiate e^(2x) again to find the second derivative, simplifying to find the final answer.
Prove: d/dx (e^(ax)^2) = 2axe^(ax)^2.
Let’s start using the chain rule: Consider y = (e^(ax))^2 = (e^(ax))^2
To differentiate, we use the chain rule: dy/dx = 2(e^(ax)) * d/dx[e^(ax)]
Since the derivative of e^(ax) is ae^(ax), dy/dx = 2(e^(ax)) * ae^(ax) = 2axe^(ax)^2 Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace e^(ax) with its derivative. As a final step, we substitute the expression back to derive the equation.
Solve: d/dx (e^ax/x).
To differentiate the function, we use the quotient rule: d/dx (e^ax/x) = (d/dx (e^ax) * x - e^ax * d/dx(x)) / x²
We will substitute d/dx (e^ax) = ae^ax and d/dx (x) = 1 = (ae^ax * x - e^ax * 1) / x² = (axe^ax - e^ax) / x²
Therefore, d/dx (e^ax/x) = (axe^ax - e^ax) / x²
In this process, we differentiate the given function using the product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.