If sin-1- x - sin-1- y-frac-pi-2- then dfrac-dy-dx- is equal to-dfrac-y-x-dfrac-y-x-dfrac-x-y-dfrac-x-y-

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Toppr Admin · Accepted Answer

Given that-sin-x2212-1x-sin-x2212-1y-x3C0-2-x2234-sin-x2212-1x-cos-x2212-1y-x21D2-y-x221A-1-x2212-x2On differentiating with respect to x- we getdydx-x2212-2x2-x221A-1-x2212-x2-x2212-xy

If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ , then $\frac{d y}{d x}$ is equal to
$- \frac{y}{x}$
$\frac{y}{x}$
$- \frac{x}{y}$
$\frac{x}{y}$

Given that,
$s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$
$∴ s i n^{- 1} x = c o s^{- 1} y$
$\Rightarrow y = \sqrt{1 - x^{2}}$
On differentiating with respect to x, we get
$\frac{d y}{d x} = \frac{- 2 x}{2 \sqrt{1 - x^{2}}} = - \frac{x}{y}$

If sin−1x+sin−1y=π2, then dydx is equal to−yxyx−xyxy

Given that,sin−1x+sin−1y=π2∴sin−1x=cos−1y⇒y=√1−x2On differentiating with respect to x, we getdydx=−2x2√1−x2=−xy

If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ , then $\frac{d y}{d x}$ is equal to
$- \frac{y}{x}$
$\frac{y}{x}$
$- \frac{x}{y}$
$\frac{x}{y}$

Given that,
$s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$
$∴ s i n^{- 1} x = c o s^{- 1} y$
$\Rightarrow y = \sqrt{1 - x^{2}}$
On differentiating with respect to x, we get
$\frac{d y}{d x} = \frac{- 2 x}{2 \sqrt{1 - x^{2}}} = - \frac{x}{y}$