Last updated on July 22nd, 2025
We use the derivative of e^(e^x), which is e^(e^x) * e^x, as a way to understand how this 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 talk about the derivative of e^(e^x) in detail.
We now understand the derivative of e^(e^x). It is commonly represented as d/dx (e^(e^x)) or (e^(e^x))', and its value is e^(e^x) * e^x. The function e^(e^x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: e^(e^x) is a composite of exponential functions.
Chain Rule: Rule for differentiating e^(e^x) since it involves a function within a function.
Exponential Derivative: The derivative of e^u is e^u * u', where u is a function of x.
The derivative of e^(e^x) can be denoted as d/dx (e^(e^x)) or (e^(e^x))'.
The formula we use to differentiate e^(e^x) is: d/dx (e^(e^x)) = e^(e^x) * e^x
The formula applies to all x, as there are no restrictions on the domain of the exponential function.
We can derive the derivative of e^(e^x) using proofs. To show this, we will use the rules of differentiation, particularly the chain rule.
There are a few methods we use to prove this, such as: Using the Chain Rule
We will now demonstrate that the differentiation of e^(e^x) results in e^(e^x) * e^x using the above-mentioned method:
To prove the differentiation of e^(e^x) using the chain rule, We consider the outer function as e^u, where u = e^x. The derivative of e^u is e^u * u'.
First, we find the derivative of the inner function, u = e^x, which is u' = e^x.
Then, we apply the chain rule: d/dx (e^(e^x)) = e^(e^x) * e^x
Therefore, the derivative of e^(e^x) is e^(e^x) * e^x. Hence, proved.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be complex.
To understand them better, think of a situation where the growth rate (first derivative) changes, and the rate of that change (second derivative) also varies. Higher-order derivatives provide deeper insights into the behavior of functions like e^(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^(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 approaches infinity, the derivative increases rapidly because e^(e^x) grows very quickly. When x is 0, the derivative of e^(e^x) = e^(e^0) * e^0 = e * 1 = e.
Students frequently make mistakes when differentiating e^(e^x). These mistakes can be resolved by understanding the proper methods. Here are a few common mistakes and ways to solve them:
Calculate the derivative of e^(e^x) * ln(x).
Here, we have f(x) = e^(e^x) * ln(x).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^(e^x) and v = ln(x).
Let’s differentiate each term, u′= d/dx (e^(e^x)) = e^(e^x) * e^x v′= d/dx (ln(x)) = 1/x
Substituting into the given equation, f'(x) = (e^(e^x) * e^x) ln(x) + e^(e^x) * (1/x)
Let’s simplify terms to get the final answer, f'(x) = e^(e^x) * e^x ln(x) + e^(e^x) / x
Thus, the derivative of the specified function is e^(e^x) * e^x ln(x) + e^(e^x) / x.
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.
XYZ Corporation is analyzing the growth of a technology, modeled by the function y = e^(e^x), where x represents time in months. Estimate the growth rate at x = 1 month.
We have y = e^(e^x) (growth model)...(1)
Now, we will differentiate the equation (1) Take the derivative of e^(e^x): dy/dx = e^(e^x) * e^x
Given x = 1 (substitute this into the derivative)
dy/dx = e^(e^1) * e^1
dy/dx = e^(e) * e
Therefore, the growth rate at x = 1 month is e^(e) * e.
We find the growth rate at x = 1 month by substituting the value of x into the derivative.
This gives us the rate at which the technology is expected to grow at that specific time.
Derive the second derivative of the function y = e^(e^x).
The first step is to find the first derivative, dy/dx = e^(e^x) * e^x...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [e^(e^x) * e^x]
Here we use the product rule, d²y/dx² = [e^(e^x) * e^x] d/dx [e^x] + d/dx [e^(e^x)] * e^x d²y/dx² = [e^(e^x) * e^x] * e^x + e^(e^x) * e^x * e^x
Simplifying, we get: d²y/dx² = e^(e^x) * (e^2x + e^x)
Therefore, the second derivative of the function y = e^(e^x) is e^(e^x) * (e^2x + e^x).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate e^(e^x) * e^x. We then simplify the terms to find the final answer.
Prove: d/dx (e^(2e^x)) = 2e^(2e^x) * e^x.
Let’s start using the chain rule: Consider y = e^(2e^x)
To differentiate, we use the chain rule: dy/dx = e^(2e^x) * d/dx [2e^x]
Since the derivative of 2e^x is 2e^x, dy/dx = e^(2e^x) * 2e^x
Substituting y = e^(2e^x), d/dx (e^(2e^x)) = 2e^(2e^x) * e^x
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace the inner function with its derivative. As a final step, we substitute y = e^(2e^x) to derive the equation.
Solve: d/dx (e^(e^x) / x).
To differentiate the function, we use the quotient rule: d/dx (e^(e^x) / x) = (d/dx (e^(e^x)) * x - e^(e^x) * d/dx(x)) / x²
We will substitute d/dx (e^(e^x)) = e^(e^x) * e^x and d/dx (x) = 1 = (e^(e^x) * e^x * x - e^(e^x)) / x² = (x * e^(e^x) * e^x - e^(e^x)) / x²
Therefore, d/dx (e^(e^x) / x) = (x * e^(e^x) * e^x - e^(e^x)) / 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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