Question

If $x=\sqrt{a^{{\sin }^{-1}t}}$ and $y=\sqrt{a^{{\cos }^{-1}t}}$ then $\frac{dy}{dx}=$

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

Answer 1

The correct option is D $-\frac{y}{x}$

Let $x=\sqrt{a^{{\sin }^{-1}t}}$ ...(1)

and $y=\sqrt{a^{{\cos }^{-1}t}}$ ...(2)

Multiplying (1) and (2), we get

$xy=\sqrt{a^{{\sin }^{-1}t}}\times \sqrt{a^{{\cos }^{-1}t}}=\sqrt{a^{{\sin }^{-1}t}\times a^{{\cos }^{-1}t}}$

$=\sqrt{a^{{\sin }^{-1}t+{\cos }^{-1}t}}=\sqrt{a^{\pi /2}}[\because {\sin }^{-1}t+{\cos }^{-1}t=\frac{\pi }{2}]$

Differentiating w.r.t $x$, we get

$x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$