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Last updated on 17 September 2025
We use the derivative of exponential functions, such as e^x, as a measuring tool for how exponential functions change in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of exponential functions in detail.
We now understand the derivative of exponential functions. The derivative of e^x is commonly represented as d/dx (\(e^x\)) or (\(e^x\))', and its value is \(e^x\).
Exponential functions have clearly defined derivatives, indicating they are differentiable across their domain.
The key concepts are mentioned below:
Exponential Function: An exponential function such as \(e^x\).
Chain Rule: Rule for differentiating complex exponential functions.
Natural Logarithm: The inverse function of the exponential function ln(x).
The derivative of e^x can be denoted as d/dx (\(e^x\)) or (\(e^x\))'. The formula we use to differentiate e^x is: d/dx (\(e^x\)) = \(e^x \)The formula applies to all x.
We can derive the derivative of \(e^x\) using proofs.
To show this, we will use the properties of exponential functions along with the rules of differentiation.
There are several methods we use to prove this, such as:
By First Principle
Using Chain Rule
Using Product Rule
We will now demonstrate that the differentiation of \(e^x\) results in e^x using the above-mentioned methods:
By First Principle
The derivative of \(e^x\) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of \(e^x\) using the first principle, we will consider f(x) = \(e^x\).
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = \(e^x\), we write f(x + h) = \(e^(x + h)\).
Substituting these into equation (1), f'(x) = limₕ→₀ [\(e^(x + h) - e^x\)] / h = limₕ→₀ [\(e^x · (e^h - 1)\)] / h = \(e^x \)· limₕ→₀ [\((e^h - 1)\) / h]
Using the limit property, limₕ→₀ (\(e^h - 1\)) / h = 1. f'(x) = \(e^x\) · 1 = \(e^x\)
Hence, proved.
Using Chain Rule
To prove the differentiation of \(e^x \)using the chain rule,
We use the formula: If y = \(e^(u)\) where u = x, then dy/dx = dy/du · du/dx = \(e^u\) · 1 = \(e^x\)
Using Product Rule
We will now prove the derivative of \(e^x\) using the product rule.
The step-by-step process is demonstrated below:
Here, we consider the function as a product of two identical exponential functions: \(e^x · e^0 = e^x · 1\)
Using the product rule formula: d/dx [u·v] = u'·v + u·v' u = \(e^x\) and v = \(e^0\) u' = d/dx (\(e^x\)) = \(e^x \)v' = d/dx (\(e^0\)) = 0 d/dx (\(e^x\)) = \(e^x · 1 + e^x · 0 = e^x\)
Thus, d/dx (\(e^x\)) =\( e^x\).
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 \(e^x\).
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^x, 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 is infinity, the derivative remains \(e^x\) because exponential functions grow indefinitely. When x is 0, the derivative of \(e^x = e^0\), which is 1.
Students frequently make mistakes when differentiating exponential functions. 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^x · x²).
Here, we have f(x) = \(e^x · x²\).
Using the product rule, f'(x) = u′v + uv′
In the given equation, u = \(e^x \)and v = x².
Let’s differentiate each term, \(u′ = d/dx (e^x) = e^x v′ = d/dx (x²) = 2x\)
Substituting into the given equation, \(f'(x) = (e^x) · (x²) + (e^x) · (2x)\)
Let’s simplify terms to get the final answer, \(f'(x) = e^x · x² + 2x · e^x\)
Thus, the derivative of the specified function is e^x · (x² + 2x).
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.
A company tracks its revenue using the function R(x) = e^(0.05x), where x represents time in months. Calculate the rate of change of revenue at x = 10 months.
We have R(x) = \(e^(0.05x)\) (rate of revenue change)...(1)
Now, we will differentiate the equation (1)
Take the derivative of \(e^(0.05x): dR/dx = 0.05e^(0.05x)\)
Given x = 10, substitute this into the derivative: \(dR/dx = 0.05e^(0.05*10) = 0.05e^0.5\)
Hence, the rate of change of revenue at 10 months is\( 0.05e^0.5\).
We find the rate of change of revenue at x = 10 by differentiating the revenue function and substituting the given value of x into the derivative.
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. Using basic differentiation rules, we differentiate \(2e^(2x)\) to find the second derivative.
Prove: d/dx (e^(3x)) = 3e^(3x).
Let’s start using the chain rule: Consider\( y = e^(3x)\)
To differentiate, we apply the chain rule:\( dy/dx = e^(3x) · d/dx (3x) = e^(3x) · 3 = 3e^(3x)\)
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation by multiplying the outer function's derivative by the inner function's derivative.
Solve: d/dx (e^x/x).
To differentiate the function, we use the quotient rule: \(d/dx (e^x/x) = (d/dx (e^x) · x - e^x · d/dx(x))/ x²\)
We will substitute \(d/dx (e^x) = e^x\) and \(d/dx (x) = 1 = (e^x · x - e^x · 1) / x² = (x · e^x - e^x) / x² = e^x(x - 1) / x²\)
Therefore, \(d/dx (e^x/x) = e^x(x - 1)/ x²\)
In this process, we differentiate the given function using the 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.
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