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Last updated on October 3, 2025
We use the derivative of e^2x^2 to understand how this exponential function changes in response to a slight change in x. Derivatives are instrumental in calculating changes in various real-life situations, such as growth rates. We will now discuss the derivative of e^2x^2 in detail.
We now understand the derivative of\( e^2x^2\). It is commonly represented as\( d/dx (e^2x^2)\) or \((e^2x^2)'\), and its value is \(4xe^2x^2\). The function\( e^2x^2\) has a clearly defined derivative, indicating it is differentiable at all points. The key concepts are mentioned below:
Exponential Function: (\(e^2x^2\) = e raised to the power of\( 2x^2\)).
Chain Rule: Rule for differentiating composite functions like \(e^2x^2\).
Constant Multiple Rule: Used when differentiating a function multiplied by a constant.
The derivative of\( e^2x^2\) can be denoted as d/dx \((e^2x^2\)) or \((e^2x^2\)').
The formula for differentiating \(e^2x^2 \)is: d/dx \((e^2x^2)\) = \(4xe^2x^2 \)
The formula applies to all x in the domain of real numbers.
We can derive the derivative of e^2x^2 using proofs. To show this, we will use differentiation rules such as the chain rule. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of \(4xe^2x^2 \)results in\( 4xe^2x^2 \)using these methods:
To prove the differentiation of e^2x^2 using the chain rule, Consider f(x) =\( 4xe^2x^2 \), and let g(u) = \(e^u\) where u = \(2x^2\).
According to the chain rule, d/dx [g(f(x))] = g'(f(x)) * f'(x).
First, find the derivative of f(x): f'(x) = d/dx (\(2x^2\)) = 4x.
Next, find the derivative of g(u): g'(u) = e^u. Substituting back, we have:
\(d/dx\) \([e^2x^2]\) = \(e^2x^2 * 4x. d/dx [e^2x^2] = 4xe^2x^2\).
Hence, proved.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can become more complex.
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 analyze functions like e^2x^2.
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^2x^2, we generally use f^(n)(x) to denote 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 any real number, the derivative is always defined because\( e^2x^2\) is defined for all real numbers. When x is 0, the derivative of \(e^2x^2 = 4x e^2x^2\), which evaluates to 0.
Students frequently make mistakes when differentiating\( e^2x^2\). 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^2x^2 * ln x)
Here, we have f(x) =\( e^2x^2 \)* ln x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = \(e^2x^2\) and v = ln x. Let’s differentiate each term, \(u′ = d/dx (e^2x^2) = 4xe^2x^2 v′ = d/dx (ln x) = 1/x \)
Substituting into the given equation,\( f'(x) = (4xe^2x^2).(ln x) + (e^2x^2).(1/x)\) Let’s simplify terms to get the final answer,\( f'(x) = 4xe^2x^2 ln x + e^2x^2 / x\) Thus, the derivative of the specified function is \(4xe^2x^2 ln x + e^2x^2 / x.\)
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.
A company tracks the growth of a colony of bacteria over time, represented by the function y = e^2x^2, where y represents the number of bacteria at time x. If x = 1 hour, measure the rate of growth of the bacteria.
We have \(y = e^2x^2\) (growth model)...(1)
Now, we will differentiate the equation (1)
Take the derivative\( e^2x^2: dy/dx = 4xe^2x^2\) Given x = 1 (substitute this into the derivative)\( dy/dx = 4(1)e^(2(1)^2) = 4e^2 \)
Therefore, the rate of growth of the bacteria at x = 1 hour is \(4e^2\).
We find the growth rate of the bacteria at x = 1 hour as \(4e^2\), which means that at one hour, the number of bacteria increases at a rate proportional to 4e^2.
Derive the second derivative of the function y = e^2x^2.
The first step is to find the first derivative, \(dy/dx = 4xe^2x^2.\)..(1)
Now we will differentiate equation (1) to get the second derivative:\( d²y/dx² = d/dx [4xe^2x^2] \)
Here we use the product rule,\( d²y/dx² = 4 * d/dx [xe^2x^2] = 4 * (e^2x^2 + 2x * 2xe^2x^2) = 4 * e^2x^2 (1 + 4x^2) \)
Therefore, the second derivative of the function \(y = e^2x^2 is 4e^2x^2 (1 + 4x^2).\)
We use the step-by-step process, starting with the first derivative. By applying the product rule, we differentiate the function further, substituting and simplifying to find the final answer.
Prove: d/dx (e^(2x^2 + 3)) = 4xe^(2x^2 + 3).
Let’s start using the chain rule: Consider y = e(2x^2 + 3)
To differentiate, we use the chain rule: \(dy/dx = e^(2x^2 + 3) * d/dx [2x^2 + 3]\)
Since the derivative of\( 2x^2 + 3 is 4x, dy/dx = e^(2x^2 + 3) * 4x\)
Substituting y = \(e^(2x^2 + 3), d/dx (e^(2x^2 + 3)) = 4xe^(2x^2 + 3)
\)
hence proved!
In this step-by-step process, we used the chain rule to differentiate the equation. We then replaced the inner function with its derivative and simplified to derive the equation.
Solve: d/dx (e^2x^2/x)
To differentiate the function, we use the quotient rule:\( d/dx (e^2x^2/x) = (d/dx (e^2x^2) * x - e^2x^2 * d/dx(x)) / x^2\)
We will substitute \(d/dx \)\((e^2x^2) = 4xe^2x^2 \) and d/dx (x) = 1 = \((4xe^2x^2 * x - e^2x^2 * 1) / x^2 = (4x^2e^2x^2 - e^2x^2) / x^2 = e^2x^2 (4x^2 - 1) / x^2\)
Therefore,\( d/dx (e^2x^2/x) = e^2x^2 (4x^2 - 1) / x^2\)
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.
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