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प्रश्न
What is \[\frac{d}{dx}\left(\sqrt{1-x^2}\right)\] in the differentiation for \[y=\sin^{-1}x\]?
पर्याय
\[\frac{2x}{2\sqrt{1-x^2}}\]
\[-\frac{2x}{2\sqrt{1-x^2}}\]
\[-2x\sqrt{1-x^2}\]
\[\frac{1}{2\sqrt{1-x^2}}\]
MCQ
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उत्तर
Using the chain rule, the derivative of \[1-x^2\] is \[-2x\]. Therefore, \[\frac{d}{dx}(\sqrt{1-x^2})=-\frac{2x}{2\sqrt{1-x^2}}\].
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