Last updated on July 23rd, 2025
We use the derivative of exp(x), which is exp(x) itself, as a tool for understanding how the exponential function changes in response to a slight change in x. Derivatives help us calculate growth or decay in real-life situations. We will now discuss the derivative of exp(x) in detail.
We now understand the derivative of exp(x). It is commonly represented as d/dx (exp(x)) or (exp(x))', and its value is exp(x). The function exp(x) has a clearly defined derivative, indicating it is differentiable for all real numbers.
The key concepts are mentioned below: Exponential Function: exp(x) = e^x.
Derivative Rule: Rule for differentiating exp(x) (the derivative of e^x is e^x).
Growth Rate: The rate at which exp(x) increases as x increases.
The derivative of exp(x) can be denoted as d/dx (exp(x)) or (exp(x))'.
The formula we use to differentiate exp(x) is: d/dx (exp(x)) = exp(x) (or) (exp(x))' = exp(x)
The formula applies to all real numbers x.
We can derive the derivative of exp(x) using various methods. To show this, we will use the rules of differentiation.
There are several methods to prove this, such as: Using the Limit Definition Using Chain Rule Using Exponential Rules We will now demonstrate that the differentiation of exp(x) results in exp(x) using the above-mentioned methods:
Using the Limit Definition The derivative of exp(x) can be proved using the limit definition, which expresses the derivative as the limit of the difference quotient.
To find the derivative of exp(x) using the limit definition, we consider f(x) = exp(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) = exp(x), we write f(x + h) = exp(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [exp(x + h) - exp(x)] / h = limₕ→₀ [exp(x) exp(h) - exp(x)] / h = exp(x) limₕ→₀ [exp(h) - 1] / h Using the fact that limₕ→₀ [exp(h) - 1] / h = 1, f'(x) = exp(x) * 1 = exp(x)
Hence, proved. Using Chain Rule To prove the differentiation of exp(x) using the chain rule, We recognize that exp(x) is its own derivative: Let u = x, and f(u) = exp(u) Then d/dx [exp(x)] = d/du [exp(u)] * du/dx Since d/du [exp(u)] = exp(u) and du/dx = 1, d/dx [exp(x)] = exp(x)
Using Exponential Rules The function f(x) = exp(x) is unique in that its derivative is the same as the function itself.
We use the exponential rule: d/dx (e^x) = e^x This directly gives us: d/dx (exp(x)) = exp(x)
When a function is differentiated multiple times, the derivatives obtained are referred to as higher-order derivatives.
Higher-order derivatives of exp(x) are straightforward since they remain the same.
For the first derivative of a function, we write f′(x), which shows the function's change or slope at a certain point.
The second derivative is derived from the first derivative, which is denoted using f′′(x) and remains exp(x). Similarly, the third derivative, f′′′(x), is also exp(x), and this pattern continues.
For the nth derivative of exp(x), we generally use fⁿ(x), and it remains exp(x), indicating the change in the rate of change.
The exponential function exp(x) is defined and differentiable for all real numbers without any exceptions. For x = 0, the derivative of exp(x) = exp(0), which is 1.
Students frequently make mistakes when differentiating exp(x).
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 (exp(x) · x^2)
Here, we have f(x) = exp(x) · x².
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = exp(x) and v = x².
Let’s differentiate each term, u′ = d/dx (exp(x)) = exp(x) v′ = d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (exp(x)) · (x²) + (exp(x)) · (2x)
Let’s simplify terms to get the final answer, f'(x) = exp(x) * x² + 2x * exp(x)
Thus, the derivative of the specified function is exp(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 models its growth with the function y = exp(x), where y represents the revenue in millions, and x is the number of years since inception. If x = 2 years, calculate the growth rate of the company's revenue.
We have y = exp(x) (growth model)...(1) Now, we will differentiate the equation (1)
Take the derivative exp(x): dy/dx = exp(x) Given x = 2 (substitute this into the derivative) exp(2) = e²
Hence, the growth rate of the company's revenue at x = 2 years is e².
We find the growth rate of the company's revenue at x = 2 years as e², indicating that at this time, the revenue is increasing at a rate equal to e² times the current revenue.
Derive the second derivative of the function y = exp(x).
The first step is to find the first derivative, dy/dx = exp(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [exp(x)]
Since the derivative of exp(x) is exp(x), d²y/dx² = exp(x)
Therefore, the second derivative of the function y = exp(x) is exp(x).
We use a step-by-step process, where we start with the first derivative. Recognizing that the derivative of exp(x) is itself, the second derivative remains exp(x).
Prove: d/dx (exp(2x)) = 2 exp(2x).
Let’s start using the chain rule: Consider y = exp(2x)
To differentiate, we apply the chain rule: dy/dx = d/dx [exp(2x)] = exp(2x) · d/dx (2x) Since d/dx (2x) = 2, dy/dx = 2 exp(2x) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Recognizing the inner function 2x, we multiply the result by its derivative to derive the equation.
Solve: d/dx (exp(x)/x)
To differentiate the function, we use the quotient rule: d/dx (exp(x)/x) = (d/dx (exp(x)) · x - exp(x) · d/dx(x)) / x² We will substitute d/dx (exp(x)) = exp(x) and d/dx (x) = 1 (exp(x) · x - exp(x) · 1) / x² = (x exp(x) - exp(x)) / x² = exp(x) (x - 1) / x² Therefore, d/dx (exp(x)/x) = exp(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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