Question

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

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

Answer 1

Answer: (C)

Given, $x=\sqrt{a^{{\sin }^{-1}t}}$
and $y=\sqrt{a^{{\cos }^{-1}t}}$
On multiplying Eqs. (i) and (ii), we get
$xy=\sqrt{a^{{\sin }^{-1}t}}\times \sqrt{a^{{\cos }^{-1}t}}$
$\Rightarrow xy=\sqrt{a^{{\sin }^{-1}t}\cdot a^{{\cos }^{-1}t}}$
$=\sqrt{a^{\left({\sin }^{-1}t+{\cos }^{-1}t\right)}}$
$\left(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\right)$
$\Rightarrow xy=\sqrt{a^{\pi /12}}$
On differentiating w.r.t. $x$, we get
$x\frac{dy}{dx}+y=0$
$\Rightarrow x\frac{dy}{dx}=-y$
$\Rightarrow \frac{dy}{dx}=\frac{-y}{x}$
$\therefore \frac{d}{dx}$ (constant $)=0$