Question

Derivative of sin(x+y)

Answer 1

`(dy)/(dx) = ( \cos(x + y))/([1 - \cos(x + y)] )`

Explanation:

Let y = sin(x+y)

Differentiating with respect to x on both sides.

`(dy)/(dx) = (d)/(dx) \sin(x + y) \\ \\ (dy)/(dx) = \cos(x + y)(d)/(dx) (x + y) \\ \\ (dy)/(dx) = \cos(x + y)(1 + (dy)/(dx)) \\ \\ (dy)/(dx) = \cos(x + y) +\cos(x + y) (dy)/(dx) \\ \\ (dy)/(dx) - \cos(x + y) (dy)/(dx) = \cos(x + y) \\ \\ (dy)/(dx) [1 - \cos(x + y)] = \cos(x + y) \\ \\ \purple {\bold {(dy)/(dx) = ( \cos(x + y))/([1 - \cos(x + y)] )}}`