Derivative of √x
The square root looks like it needs its own rule — until you write it as x^(1/2) and watch the ordinary power rule handle it.
Step-by-step solution
√x = x1/2. Fractional exponents are still just exponents.
d/dx[x1/2] = (1/2)·x1/2 − 1 = (1/2)·x−1/2.
x−1/2 = 1/√x, so f′(x) = 1/(2√x).
Why it works
The formula quietly predicts the square root's most famous behavior: as x → 0⁺ the slope 1/(2√x) blows up to infinity, which is why the graph leaves the origin vertically. And as x grows, the slope decays toward zero — the root keeps rising forever but ever more slowly. One derivative, both ends of the story.
This is also the standard demonstration that the power rule isn't just for whole numbers. On positive inputs, it works for any real exponent — halves, thirds, negatives, even irrational powers like x^π — which collapses square roots, cube roots, and reciprocals into a single rule instead of three separate formulas.
Where the derivative exists
The real square-root function is defined for x ≥ 0, but its derivative 1/(2√x) is finite only for x > 0. At zero, the right-hand difference quotient is √h/h = 1/√h, which tends to positive infinity as h approaches zero. This gives a vertical tangent, not a finite derivative at zero.
For √g(x), the chain-rule formula is g′(x)/(2√g(x)) at points where g is differentiable and g(x) > 0. Points where g(x) = 0 require separate analysis. For example, √(x²) = |x| has derivative −1 for x < 0 and 1 for x > 0, but no derivative at zero.
For √(x² + 1), the inner expression is always positive, so the derivative x/√(x² + 1) exists for all real x. The power rule for a general real exponent is safely applied on positive inputs; other domains depend on the exponent.
Common mistakes
- Answering 1/√x and losing the factor of 2 in the denominator.
- Computing 1/2 − 1 as +1/2 instead of −1/2 — the sign of the new exponent is where most slips happen.
- Forgetting the chain rule on √(inner): d/dx[√(x²+1)] = x/√(x²+1), not 1/(2√(x²+1)).
Practice problems
Differentiate 6√x
Answer: 3/√x.
Differentiate √x + 1/√x
Answer: 1/(2√x) − 1/(2x√x) — write the second term as x^(−1/2) first.
Find the slope of √x at x = 25
Answer: 1/(2·5) = 1/10.