Question

Sinxy+cosxy=1 show that dy/dx=-y/x

Answer 1

$\huge\underline\mathfrak\red{Answer}$

$\sin(x.y) + \cos(x.y) = 1 \\ \\ \sin(x.y) = 1 - \cos(x.y ) \\ \\ take \: square \: of \: both \: side \\ \\ {sin}^{2} (x.y) = 1 + {cos}^{2} (x.y) - 2 \cos(x.y) \\ \\ convert \: whole \: equation \: in \: \cos(x.y) \\ \\ 1 - {cos}^{2} (x.y) = 1 + {cos}^{2} (x.y) - 2 \cos(x.y) \\ \\ 2 {cos}^{2} (x.y) - 2 \cos(x.y) = 0 \\ \\2 \cos(x.y) ( \cos(x.y) - 1) = 0 \\ \\ from \: this \: we \: get \\ \\ 2 \cos(x.y) = 0 \\ \\ \cos(x.y) = 0 \\ \\ \cos(x.y) = \cos( \frac{\pi}{2} ) \\ \\ x.y = \frac{\pi}{2} \\ \\ \frac{d(x.y)}{dx} = \frac{d( \frac{\pi}{2} )}{dx} \\ \\ x. \frac{dy}{dx} + y. \frac{dx}{dx} = 0 \\ \\ x. \frac{dy}{dx} = - y \\ \\ \frac{dy}{dx} = \frac{ - y}{x}$