Question

If $x=\sqrt{a^{si{n}^{-1}t}}$ and $y=\sqrt{a^{co{s}^{-1}t}}$, 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

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

Given, $x=\sqrt{a^{si{n}^{-1}t}}$ .... $\left(i\right)$
and $y=\sqrt{a^{co{s}^{-1}t}}$ ..... $\left(ii\right)$
On multiplying Eqs. $\left(i\right)$ and $\left(ii\right),$ we get
$xy=\sqrt{a^{si{n}^{-1}t}}\times \sqrt{a^{co{s}^{-1}t}}$
$\Rightarrow xy=\sqrt{a^{si{n}^{-1}t}.a^{co{s}^{-1}t}}$
$=\sqrt{a^{(si{n}^{-1}t+co{s}^{-1}t)}}$
$(\because si{n}^{-1}t+co{s}^{-1}t=\frac{\pi }{2})$
$\Rightarrow xy=\sqrt{a^{\pi /2}}$
On differentating w.r.t. x, we get
$x\frac{dy}{dx}+y=0$ .... $[\because \frac{d}{dx}(constant)=0]$

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

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