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Worked solution

An inverse sine derivative with two branches

Find ddx[sin⁡−11−x2]\dfrac d{dx}[\sin^{-1}\sqrt{1-x^2}].

[N99/I/12]

CJC · Discussion · Question 6

Answer

−x∣x∣1−x2-\dfrac{x}{|x|\sqrt{1-x^2}} for 0<∣x∣<10<|x|<1: positive branch −1/1−x2-1/\sqrt{1-x^2}; negative branch 1/1−x21/\sqrt{1-x^2}. Not differentiable at x=0x=0.

Chain rule

Apply the inverse-sine and square-root chain rules: 11−(1−x2)2⋅12(1−x2)−1/2(−2x)\dfrac1{\sqrt{1-(\sqrt{1-x^2})^2}}\cdot\dfrac12(1-x^2)^{-1/2}(-2x). This is −x1−x2⋅1x2=−x∣x∣1−x2-\dfrac{x}{\sqrt{1-x^2}}\cdot\dfrac1{\sqrt{x^2}}=-\dfrac{x}{|x|\sqrt{1-x^2}} for 0<∣x∣<10<|x|<1.

How did this part go?

Positive inputs

For 0<x<10<x<1, ∣x∣=x|x|=x, so the derivative is −1/1−x2-1/\sqrt{1-x^2}.

How did this part go?

Negative inputs

For −1<x<0-1<x<0, ∣x∣=−x|x|=-x, so the derivative is 1/1−x21/\sqrt{1-x^2}. The source highlights x2=∣x∣\sqrt{x^2}=|x|. As an additional domain check, the original function is arccos⁡∣x∣\arccos|x| on [−1,1][-1,1]; its left derivative at zero is +1 and its right derivative is −1, so it is not differentiable there.

How did this part go?

Common mistakes

Cancelling x against |x| without splitting signs gives the wrong negative-input derivative.

Solution notes

The source gives separate x>0 and x<0 branches and the absolute-value identity. The precise open derivative intervals and the nondifferentiability at zero are authored supplements; finite ordinary derivatives are not asserted at the domain endpoints ±1.

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