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Last updated on September 15, 2025
The derivative of e to the x squared, denoted as d/dx (e^(x²)), is essential for understanding how the function changes as x varies. Derivatives are powerful tools in various applications, from calculating rates of change to modeling real-life situations. We will now explore the derivative of e^(x²) in detail.
To understand the derivative of e to the x squared, we express it as d/dx (e(x²)) or (e(x²))'.
The derivative is 2x * e(x²), indicating that the function is differentiable for all real numbers.
The key concepts include:
Exponential Function: e(x²) is an exponential function of x².
Chain Rule: A differentiation technique necessary because the exponent is a function of x.
Exponential Growth: The function exhibits exponential growth as x increases.
The derivative of e(x²) is expressed as d/dx (e(x²)) or (e(x²))'.
The formula used to differentiate e(x²) is: d/dx (e(x²)) = 2x * e(x²)
The formula is valid for all real numbers x.
We can derive the derivative of e(x²) using different methods, such as:
Using the Chain Rule
Using the First Principle We will demonstrate how the differentiation of e(x²) results in 2x * e(x²) using these methods:
Using the Chain Rule
To differentiate e^(x²) using the chain rule, we consider the exponent as a function of x. Let u = x², then f(u) = eu and u' = 2x.
Therefore, by the chain rule: d/dx (e(x²)) = d/du (eu) * u' = e(x²) * 2x = 2x * e(x²)
By First Principle We can also prove the derivative using the First Principle, which expresses the derivative as the limit of the difference quotient.
Consider f(x) = e(x²). Its derivative is: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h
Substitute f(x) = e(x²), so f(x + h) = e^((x + h)²): f'(x) = limₕ→₀ [e((x + h)²) - e(x²)] / h
To solve this, we expand (x + h)² and apply the limit.
Ultimately, this results in the derivative 2x * e(x²), aligning with the chain rule proof.
Higher-order derivatives are obtained by differentiating a function multiple times.
These derivatives provide insights into the behavior of the function, similar to how speed and acceleration describe a car's motion.
For e(x²), higher-order derivatives can be expressed as follows:
The first derivative is 2x * e(x²), representing the rate of change of the function.
The second derivative is derived from the first derivative, denoted f′′(x).
The third derivative, f′′′(x), follows from the second derivative, and this pattern continues.
For the nth derivative of e(x²), we use fⁿ(x) to represent the nth derivative, indicating the change in the rate of change.
When x = 0, the derivative is 0 because e(0²) = 1 and 2x = 0. For large positive or negative x, the derivative grows rapidly due to the exponential function's nature.
Students often make errors when differentiating e(x²). These mistakes can be resolved by understanding the correct procedures. Here are some common mistakes and solutions:
Calculate the derivative of e^(x²) * sin(x).
Let f(x) = e(x²) * sin(x).
Using the product rule, f'(x) = u′v + uv′.
Here, u = e(x²) and v = sin(x).
Differentiate each term: u′ = d/dx (e(x²)) = 2x * e(x²)
v′ = d/dx (sin(x)) = cos(x)
Substitute into the equation: f'(x) = (2x * e(x²)) * sin(x) + e(x²) * cos(x)
Simplifying gives: f'(x) = 2x * e(x²) * sin(x) + e(x²) * cos(x)
We find the derivative by breaking the function into two parts and applying the product rule. Each part is differentiated, then combined to give the final result.
In a physics experiment, the distance traveled by a particle is modeled by y = e^(x²), where x is time. Find the rate of change of distance at x = 1 second.
Given y = e(x²), differentiate to find the rate of change:
dy/dx = 2x * e(x²)
Substitute x = 1: dy/dx = 2 * 1 * e(1²) = 2e
Hence, at x = 1 second, the rate of change of distance is 2e.
The rate of change at x = 1 is found by substituting into the derivative. This gives the instantaneous rate of change of distance at that point.
Derive the second derivative of y = e^(x²).
First, find the first derivative:
dy/dx = 2x * e(x²)
Now differentiate again to find the second derivative:
d²y/dx² = d/dx [2x * e(x²)]
Apply the product rule: = 2 * e(x²) + 2x * (2x * e(x²)) = 2 * e(x²) + 4x² * e(x²)
Simplify to: d²y/dx² = 2e(x²) * (1 + 2x²)
We derive the second derivative by applying the product rule to the first derivative. This involves differentiating each component and simplifying.
Prove: d/dx ((e^(x²))²) = 4x * e^(2x²).
Let y = (e(x²))² = e(2x²).
Differentiate using the chain rule:
dy/dx = d/dx [e(2x²)] = 2x * e(2x²) * 2 = 4x * e(2x²)
Hence proved.
Using the chain rule, we differentiate e(2x²) by considering the inner function 2x². The derivative is verified through substitution and simplification.
Solve: d/dx (e^(x²)/x).
Use the quotient rule to differentiate:
d/dx (e(x²)/x) = [(d/dx (e^(x²)) * x) - (e(x²) * 1)] / x²
Substitute d/dx (e(x²)) = 2x * e(x²): = [(2x * e(x²) * x) - e(x²)] / x² = (2x² * e(x²) - e(x²)) / x² = e(x²) * (2x² - 1) / x²
The problem is solved using the quotient rule, applying derivatives to numerator and denominator separately, and simplifying the expression for 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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