How do you find the derivative of #y = sin(x+y)#?

1 Answer
Jul 28, 2016

#dy/dx= cos(x+y)/{1-cos(x+y)}#

Explanation:

You simply differentiate both sides with respect to #x#. The left side would simply give you #dy/dx#. For the right side, however, you must make use of the chain rule for derivatives of composite functions (functions of functions). Thus

#d/dx (sin(x+y)) = cos(x+y) xx d/dx (x+y) = cos(x+y) (1+dy/dx)#

Thus, we get

#dy/dx = cos(x+y) (1+dy/dx)#

We can easily solve this for the quantity #dy/dx#:

#(1-(cos(x+y)) dy/dx = cos(x+y) implies dy/dx= cos(x+y)/{1-cos(x+y)}#