Question

If ${\sin }^{1}x+{\sin }^{-1}y=\frac{\pi }{2}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{x}{y}$
• B. $-\frac{x}{y}$
• C. $\frac{y}{x}$
• D. $-\frac{y}{x}$

Answer 1

Answer: (B)

$-\frac{x}{y}$

Given, ${\sin }^{1}x+{\sin }^{-1}y=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}y$

$\Rightarrow {\sin }^{-1}x={\cos }^{-1}y$

$\Rightarrow {\cos }^{-1}\sqrt{1-x^2}={\cos }^{-1}y$

$\Rightarrow \sqrt{1-x^2}=y$

On differentiating both sides w.r.t. $x$, we get

$\frac{1}{2\sqrt{1-x^2}}\times (-2x)=\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=-\frac{x}{\sqrt{1-x^2}}=-\frac{x}{y}$.