Answer:
[tex] \frac{dy}{dx} = \frac{ \cos(x + y)}{[1 - \cos(x + y)] } [/tex]
Step-by-step explanation:
Let y = sin(x+y)
Differentiating with respect to x on both sides.
[tex] \frac{dy}{dx} = \frac{d}{dx} \sin(x + y) \\ \\ \frac{dy}{dx} = \cos(x + y)\frac{d}{dx} (x + y) \\ \\ \frac{dy}{dx} = \cos(x + y)(1 + \frac{dy}{dx}) \\ \\ \frac{dy}{dx} = \cos(x + y) +\cos(x + y) \frac{dy}{dx} \\ \\ \frac{dy}{dx} - \cos(x + y) \frac{dy}{dx} = \cos(x + y) \\ \\ \frac{dy}{dx} [1 - \cos(x + y)] = \cos(x + y) \\ \\ \purple {\bold {\frac{dy}{dx} = \frac{ \cos(x + y)}{[1 - \cos(x + y)] }}} [/tex]
