Question

sin-1x + sin-1y = pi/2... then find dy/dx

Answer 1

Hii hope it helps you

Answer 2

Answer:

$-\frac{x}{y}$

Step-by-step explanation:

$\frac{d(sin^{-1} x)}{dx} =\frac{1}{\sqrt{1-x^{2} } }$

⇒$sin^{-1} x-sin^{-1} y=\frac{\pi}{2}$

differentiating with respect to x.

$\frac{1}{\sqrt{1-x^{2} } }*\frac{dx}{dx} -\frac{1}{\sqrt{1-y^{2} } }*\frac{dy}{dx} =0$

$\frac{dy}{dx} =-\frac{\sqrt{1-y^{2} } }{\sqrt{1-x^2} }$ --------1

$as, sin^{-1}x-sin^{-1}y=\frac{\pi}{2}$

$sin^{-1}x=\frac{\pi}{2}-sin^{-1}y$

using property of inverse,$\frac{\pi}{2}-sin^{-1}y=cos^{-1}y$

⇒$sin^{-1}x=cos^{-1}y$

and $sin^{-1}x=cos^{-1}\sqrt{1-x^{2}}$

⇒ $cos^{-1}\sqrt{1-x^{2}}=cos^{-1}y$

⇒$\sqrt{1-x^{2}} =y$ ⇒ $\sqrt{1-y^{2}} =x$

threfore,$\frac{dy}{dx} =-\frac{\sqrt{1-y^{2} } }{\sqrt{1-x^2} }$  becomes,

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