Question

If $y=\sqrt{\frac{1-{\sin }^{-1}x}{1+{\sin }^{-1}x}}$ then $y^{\prime }\left(0\right)$ is equal to

• A. $1$
• B. $1/2$
• C. $-1$
• D. $\sqrt{2}/3$

Answer 1

Answer: (C)

$-1$

Put $x=\sin \left(\cos \theta \right)$ so $y=\tan \frac{\theta }{2}$

Thus

$\frac{dy}{dx}=\left(\frac{1}{2}{\sec }^2\frac{\theta }{2}\right)(-\frac{1}{\sin \theta \left(\cos \theta \right)})$

$\Rightarrow y^{\prime }\left(0\right)=-1$