Question

What is \[\frac{d}{dx}\left(\sqrt{1-x^2}\right)\] in the differentiation for \[y=\sin^{-1}x\]?

Options

• \[\frac{2x}{2\sqrt{1-x^2}}\]

• \[-\frac{2x}{2\sqrt{1-x^2}}\]

• \[-2x\sqrt{1-x^2}\]

• \[\frac{1}{2\sqrt{1-x^2}}\]

Answer 1

Using the chain rule, the derivative of \[1-x^2\] is \[-2x\]. Therefore, \[\frac{d}{dx}(\sqrt{1-x^2})=-\frac{2x}{2\sqrt{1-x^2}}\].