Question

$\sin(x) + \sin(y) = \tan(xy)$
differentiation by dy/dx​

Answer 1

$Given \: \: Question \: Is \: \\ \\ \sin(x) + \sin(y) = \tan(xy) \\ \\ Differentiate \: \: Both \: Sides \: With \: \\ respect \: to \: x \: we \: have \\ \\ \cos(x) + \cos(y) \frac{dy}{dx} = \sec {}^{2} (xy) (y + x \frac{dy}{dx} ) \\ \\ \cos( y) \frac{dy}{dx} - \sec {}^{2} (xy) x \frac{dy}{dx} = y \sec {}^{2} (xy) - \cos(x) \\ \\ \frac{dy}{dx} ( \cos(y) - x\sec {}^{2} (xy)) = y \sec {}^{2} (xy) - \cos(x) \\ \\ \frac{dy}{dx} = \frac{y \sec {}^{2} (xy) - \cos(x) }{ \cos(y) - x \sec {}^{2} (xy) } \\ \\ Note \: \\ \\ 1) \: \: \: if \: \: y = uv \: \: \: then \: its \: Differentiation \: is \\ \\ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$