AP Calculus AB and BC

Derivative of sqrt(1-x^2): Answer, Proof, Mistakes

The derivative of the square root of 1 - x^2 is -x divided by the square root of 1 - x^2, valid for -1 < x < 1. In prime notation, if f(x) = sqrt(1 - x^2) then f'(x) = -x/sqrt(1 - x^2). Rewrite the root as a 1/2 power and apply the chain rule with inner derivative -2x.

ddx[1−x2]=−x1−x2\frac{d}{dx}\left[\sqrt{1-x^2}\right] = \frac{-x}{\sqrt{1-x^2}}

How to differentiate sqrt(1-x^2)

Write the radical as a power, 1−x2=(1−x2)12\sqrt{1-x^{2}} = \left(1-x^{2}\right)^{\frac{1}{2}}, then use the chain rule with inner function 1−x21-x^{2} and inner derivative −2x-2x.

ddx(1−x2)12=12(1−x2)−12⋅(−2x)=−x1−x2\frac{d}{dx}\left(1-x^{2}\right)^{\frac{1}{2}} = \frac{1}{2}\left(1-x^{2}\right)^{-\frac{1}{2}}\cdot(-2x) = \frac{-x}{\sqrt{1-x^{2}}}

The 12\frac{1}{2} out front and the −2x-2x from inside combine to leave a bare −x-x on top. The formula holds on −1<x<1-1<x<1; at x=±1x = \pm 1 the denominator is 00 and the tangent line is vertical.

The unit circle check

y=1−x2y = \sqrt{1-x^{2}} is the upper half of the circle x2+y2=1x^{2}+y^{2} = 1. Differentiating the circle implicitly gives 2x+2ydydx=02x+2y\frac{dy}{dx} = 0, so the slope is −xy-\frac{x}{y}.

dydx=−xy=−x1−x2\frac{dy}{dx} = -\frac{x}{y} = \frac{-x}{\sqrt{1-x^{2}}}

The two routes agree, which is a useful sanity check whenever a semicircle appears. The geometry matches as well: the slope is 00 at x=0x = 0, the top of the circle, and runs toward −∞-\infty as xx approaches 11 from the left, where the circle turns vertical.

This function shows up in related rates problems on ladders and circles, and it is the curve whose area from −1-1 to 11 gives π2\frac{\pi}{2}.

Common mistakes with the derivative of sqrt(1-x^2)

  • Answering 121−x2\frac{1}{2\sqrt{1-x^{2}}}, using the square root rule and skipping the inner derivative.
  • Losing the minus sign. The inner derivative of 1−x21-x^{2} is −2x-2x, so the numerator ends up negative.
  • Cancelling the 22 wrongly and answering −2x1−x2\frac{-2x}{\sqrt{1-x^{2}}}; the 12\frac{1}{2} from the power rule halves it.
  • Quoting a derivative at x=1x = 1 or x=−1x = -1, where the tangent is vertical and f′f' does not exist.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of 1−x2\sqrt{1-x^{2}}?

It is −x1−x2\frac{-x}{\sqrt{1-x^{2}}} for −1<x<1-1<x<1, by the chain rule on (1−x2)12\left(1-x^{2}\right)^{\frac{1}{2}}.

Why is the derivative negative for positive xx?

Because the upper semicircle falls as xx increases past 00. Algebraically the sign comes from the inner derivative −2x-2x.

Where is the derivative undefined?

At x=±1x = \pm 1, where 1−x2=0\sqrt{1-x^{2}} = 0 and the tangent is vertical. Outside [−1,1][-1,1] the function itself is undefined.