Question

If $x=\sqrt{a^{\sin ^{-1}} t}$, and $y=\sqrt{a^{\cos ^{-1}} t}$, then the value of $\frac{d y}{d x}$ 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
$x y=\sqrt{a^{\sin ^{-1} t}} \times \sqrt{a^{\cos ^{-1} t}}$
$\Rightarrow x y =\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 x y=\sqrt{a^{\pi / 12}}$
On differentiating w.r.t. $x$, we get
$x \frac{d y}{d x}+y=0 $
$\Rightarrow x \frac{d y}{d x}=-y $
$ \Rightarrow \frac{d y}{d x}=\frac{-y}{x}$
$\therefore \frac{d}{d x}$ (constant $)=0$