Question

If $y=si{n}^{-1}\frac{\sqrt{(1+x)}+\sqrt{(1-x)}}{2}$, then $\frac{dy}{dx}=$

• A. $\frac{1}{\sqrt{(1-x^2)}}$
• B. $-\frac{1}{\sqrt{(1-x^2)}}$
• C. $-\frac{1}{2\sqrt{(1-x^2)}}$
• D. None of these

Answer 1

The correct option is C $-\frac{1}{2\sqrt{(1-x^2)}}$

$y=si{n}^{-1}\frac{\sqrt{1+x}+\sqrt{1-x}}{2}Putx=cos\theta \therefore y=si{n}^{-1}\frac{\sqrt{1+cos\theta }+\sqrt{1-cos\theta }}{2}=si{n}^{-1}\frac{\sqrt{2co{s}^2\left(\frac{\theta }{2}\right)}+\sqrt{2si{n}^2\left(\frac{\theta }{2}\right)}}{2}=si{n}^{-1}\frac{\sqrt{2}(cos\frac{\theta }{2}+sin\frac{\theta }{2})}{2}=si{n}^{-1}(\frac{1}{\sqrt{2}}cos\frac{\theta }{2}+\frac{1}{\sqrt{2}}sin\frac{\theta }{2})=si{n}^{-1}(sin\frac{\pi }{4}cos\frac{\theta }{2}+cos\frac{\pi }{4}sin\frac{\theta }{2})=si{n}^{-1}sin(\frac{\pi }{4}+\frac{\theta }{2})=\frac{\pi }{4}+\frac{\theta }{2}\therefore y=\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}x\therefore \frac{dy}{dx}=\frac{1}{2}\frac{-1}{\sqrt{1-x^2}}=\frac{-1}{2\sqrt{1-x^2}}$