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Last updated on September 15, 2025
We use the derivative of e^-x^3 as a measuring tool for how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of e^-x^3 in detail.
We now understand the derivative of e^-x^3. It is commonly represented as d/dx (e^-x^3) or (e^-x^3)', and its value is -3x²e^-x^3. The function e^-x^3 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Exponential Function: (e^u, where u = -x^3).
Chain Rule: Rule for differentiating e^-x^3 since it involves a composite function.
Power Rule: Used in conjunction with the chain rule to differentiate -x^3.
The derivative of e^-x^3 can be denoted as d/dx (e^-x^3) or (e^-x^3)'.
The formula we use to differentiate e^-x^3 is: d/dx (e^-x^3) = -3x²e^-x^3
The formula applies to all real x.
We can derive the derivative of e^-x^3 using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:
We will demonstrate that the differentiation of e^-x^3 results in -3x²e^-x^3 using these methods:
To prove the differentiation of e^-x^3 using the chain rule, Consider f(x) = e^u where u = -x^3. Using the chain rule, d/dx [e^u] = e^u · du/dx.
Let's differentiate u = -x^3: du/dx = -3x².
Substituting back, d/dx (e^-x^3) = e^-x^3 · (-3x²) = -3x²e^-x^3.
Hence, proved.
The product rule isn't directly applicable to e^-x^3 as it is a single function, not a product of two functions, so we rely on the chain rule here.
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^3.
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^3, we generally use f^n(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 = 0, the derivative of e^-x^3 is -3(0)²e^0, which is 0. For large positive or negative x, the exponential term e^-x^3 rapidly approaches 0, making the derivative approach 0 as well.
Students frequently make mistakes when differentiating e^-x^3. 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^3 · x^2)
Here, we have f(x) = e^-x^3 · x².
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^-x^3 and v = x².
Let's differentiate each term, u′ = d/dx (e^-x^3) = -3x²e^-x^3 v′ = d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (-3x²e^-x^3) · x² + e^-x^3 · 2x
Let's simplify terms to get the final answer, f'(x) = -3x^4e^-x^3 + 2xe^-x^3
Thus, the derivative of the specified function is -3x^4e^-x^3 + 2xe^-x^3.
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.
An engineering company is modeling the cooling of a material using the function T(x) = e^-x^3 where T represents the temperature. If x = 1, find the rate of change of temperature.
We have T(x) = e^-x^3 (temperature model)...(1)
Now, we will differentiate the equation (1) Take the derivative of e^-x^3: dT/dx = -3x²e^-x^3
Given x = 1 (substitute this into the derivative)
dT/dx = -3(1)²e^-1^3 = -3e^-1
Hence, the rate of change of temperature at x = 1 is -3/e.
We find the rate of change of temperature at x = 1 as -3/e, which means that the temperature decreases at this rate as x increases.
Derive the second derivative of the function f(x) = e^-x^3.
The first step is to find the first derivative, f'(x) = -3x²e^-x^3...(1)
Now we will differentiate equation (1) to get the second derivative: f''(x) = d/dx [-3x²e^-x^3]
Using the product rule, f''(x) = d/dx [-3x²] · e^-x^3 + (-3x²) · d/dx [e^-x^3] = [-6x · e^-x^3] + (-3x²) · [-3x²e^-x^3] = -6xe^-x^3 + 9x^4e^-x^3
Therefore, the second derivative of the function f(x) = e^-x^3 is -6xe^-x^3 + 9x^4e^-x^3.
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate -3x² and e^-x^3 separately. We then substitute and simplify the terms to find the final answer.
Prove: d/dx (x^3e^-x^3) = (3x² - x^5)e^-x^3.
Let's start using the product rule: Consider y = x^3e^-x^3.
To differentiate, we use the product rule: dy/dx = d/dx [x^3] · e^-x^3 + x^3 · d/dx [e^-x^3] = 3x²e^-x^3 + x^3 · (-3x²e^-x^3) = 3x²e^-x^3 - 3x^5e^-x^3 = (3x² - x^5)e^-x^3
Hence proved.
In this step-by-step process, we used the product rule to differentiate the equation. We differentiate each term separately and simplify to derive the equation.
Solve: d/dx (e^-x^3/x).
To differentiate the function, we use the quotient rule: d/dx (e^-x^3/x) = (d/dx (e^-x^3) · x - e^-x^3 · d/dx(x))/x²
We will substitute d/dx (e^-x^3) = -3x²e^-x^3 and d/dx (x) = 1 = (-3x²e^-x^3 · x - e^-x^3 · 1) / x² = (-3x³e^-x^3 - e^-x^3) / x² = -e^-x^3(3x³ + 1)/x²
Therefore, d/dx (e^-x^3/x) = -e^-x^3(3x³ + 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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