Question

Find dy/dx of x^3+y^3=sin(x+y)

Answer 1

Answer:

Find the attachement

Step-by-step explanation:

Hi

Hope this helps you

Answer 2

$HEYA \: \\ \\ x {}^{3} + y {}^{3} = \sin(x + y) \\ Differentiate \: both \: sides \: with \: \\ respect \: to \: x \: we \: have \\ \\ 3x {}^{2} + 3y {}^{2} \: \frac{dy}{dx} = \cos(x + y) (1 + \frac{dy}{dx}) \\ \\ 3y {}^{2} \frac{dy}{dx} - \cos(x + y) \frac{dy}{dx} = \cos(x + y) - 3x {}^{2} \\ \\ \frac{dy}{dx} (3y {}^{2} - \cos(x + y) ) = \cos(x + y) - 3x {}^{2} \\ \\ \frac{dy}{dx} = \frac{ \cos(x + y) - 3x {}^{2} }{3y {}^{2} - \cos(x + y) }$