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Last updated on September 15, 2025
We use the derivative of x|x|, which is |x| + x * (d/dx |x|), 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 x|x| in detail.
We now understand the derivative of x|x|. It is represented as d/dx (x|x|) and its value is |x| + x * (d/dx |x|). The function x|x| has a clearly defined derivative, indicating it is differentiable except at x = 0. The key concepts are mentioned below:
Absolute Value Function: |x|.
Piecewise Definition: |x| is defined as x for x ≥ 0 and -x for x < 0.
Derivative of Absolute Value: d/dx |x| = x/|x| for x ≠ 0.
The derivative of x|x| can be denoted as d/dx (x|x|).
The formula we use to differentiate x|x| is: d/dx (x|x|) = |x| + x * (d/dx |x|) = 2x for x > 0 and 0 for x < 0.
The formula applies to all x except x = 0, where |x| changes from negative to positive.
We can derive the derivative of x|x| using proofs. To show this, we will use the piecewise definition of |x| along with the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of x|x| results in 2x for x > 0 and 0 for x < 0 using the above-mentioned methods:
The derivative of x|x| can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of x|x| using the first principle, we will consider f(x) = x|x|. Its derivative can be expressed as the following limit. f'(x) = lim(h→0) [f(x + h) - f(x)] / h
Given that f(x) = x|x|, we write f(x + h) = (x + h)|x + h|.
Substituting these into the equation, f'(x) = lim(h→0) [(x + h)|x + h| - x|x|] / h For x > 0, |x| = x, hence |x + h| = x + h
when h is small and positive. f'(x) = lim(h→0) [(x + h)(x + h) - x²] / h = lim(h→0) [x² + 2xh + h² - x²] / h = lim(h→0) [2xh + h²] / h = lim(h→0) [2x + h] = 2x For x < 0, |x| = -x,
hence |x + h| = -(x + h) when h is small. f'(x) = lim(h→0) [-(x + h)(x + h) - (-x)x] / h = lim(h→0) [-x² - 2xh - h² + x²] / h = lim(h→0) [-2xh - h²] / h = lim(h→0) [-2x - h] = 0 Hence, proved.
To prove the differentiation of x|x| using the chain rule, We use the formula: x|x| = x * (x for x ≥ 0 and -x for x < 0) Consider f(x) = x and g(x) = |x|
By product rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) Let’s substitute f(x) = x and g(x) = |x| d/dx (x|x|) = 1 · |x| + x · (d/dx |x|) For x > 0, |x| = x and d/dx |x| = 1,
hence d/dx (x|x|) = x + x · 1 = 2x For x < 0, |x| = -x and d/dx |x| = -1, hence d/dx (x|x|) = -x + x · (-1) = 0
We will now prove the derivative of x|x| using piecewise differentiation: Here, we use the formula, x|x| = { x², for x ≥ 0 -x², for x < 0 }
Differentiate each piece: For x ≥ 0: d/dx (x²) = 2x For x < 0: d/dx (-x²) = 0 Thus, d/dx (x|x|) = { 2x, for x > 0 0, for x < 0 }
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 x|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 x|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 = 0, the derivative is undefined because |x| is non-differentiable at 0. When x > 0, the derivative of x|x| = 2x. When x < 0, the derivative of x|x| = 0.
Students frequently make mistakes when differentiating x|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 (x|x| + x²)
Here, we have f(x) = x|x| + x².
Using the sum rule, f'(x) = d/dx (x|x|) + d/dx (x²) For x > 0, d/dx (x|x|) = 2x and d/dx (x²) = 2x, f'(x) = 2x + 2x = 4x For x < 0, d/dx (x|x|) = 0 and d/dx (x²) = 2x, f'(x) = 0 + 2x = 2x
Thus, the derivative of the specified function is 4x for x > 0 and 2x for x < 0.
We find the derivative of the given function by dividing it into parts. The first step is finding the derivative of each part and then combining them using the sum rule to get the final result.
A car's speed is modeled by the function v = x|x| where v represents velocity and x is time in seconds. If x = 2 seconds, find the acceleration of the car.
We have v = x|x| (velocity of the car)...(1)
Now, we will differentiate the equation (1) Take the derivative of x|x|: dv/dx = |x| + x * (d/dx |x|) For x > 0, |x| = x and (d/dx |x|) = 1, dv/dx = x + x * 1 = 2x
Given x = 2 seconds, substitute this into the derivative, dv/dx = 2 * 2 = 4
Hence, we get the acceleration of the car at time x = 2 seconds as 4 m/s².
We find the acceleration of the car at x = 2 seconds as 4 m/s², which means that at this point, the velocity of the car is increasing at a rate of 4 meters per second squared.
Derive the second derivative of the function y = x|x|.
The first step is to find the first derivative, dy/dx = |x| + x * (d/dx |x|) For x > 0, dy/dx = 2x For x < 0, dy/dx = 0
Now we will differentiate to get the second derivative: d²y/dx² = d/dx (2x) for x > 0 = 2 d²y/dx² = d/dx (0) for x < 0 = 0
Therefore, the second derivative of the function y = x|x| is 2 for x > 0 and 0 for x < 0.
We use the step-by-step process, where we start with the first derivative. Then, we differentiate based on the piecewise definition to find the second derivative.
Prove: d/dx ((x|x|)²) = 4x² for x > 0, 0 for x < 0
Let’s start using the chain rule: Consider y = (x|x|)²
To differentiate, we use the chain rule: dy/dx = 2(x|x|) * d/dx (x|x|) For x > 0, d/dx (x|x|) = 2x dy/dx = 2(x²) * 2x = 4x³ For x < 0, d/dx (x|x|) = 0 dy/dx = 2((-x)²) * 0 = 0
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. We replace d/dx (x|x|) with its piecewise derivative and simplify the result.
Solve: d/dx (x|x|/x)
To differentiate the function, we use simplification: d/dx (x|x|/x) = d/dx (|x|) For x > 0, |x| = x d/dx (|x|) = 1 For x < 0, |x| = -x d/dx (|x|) = -1
Therefore, d/dx (x|x|/x) = 1 for x > 0, -1 for x < 0
In this process, we simplify the given function to |x| and differentiate based on the piecewise definition of the absolute value.
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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