Derivative of x|x|
2026-02-28 23:50 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-x%7Cx%7C

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:

  1. By First Principle
  2. Using Chain Rule
  3. Using Piecewise Differentiation

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:

By First Principle

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.

Using Chain Rule

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

Using Piecewise Differentiation

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 }