Question

What is the derivative of `x = tan (x+y)`?

Answer 1

Consider `y` as a function of `x` and get (deriving both sides with respect to `x`):

`1=[1/(cos^2(x+y))]*[1+(dy)/(dx)]` (Chain Rule for `tan`)

Rearranging:

`1+(dy)/(dx)=cos^2(x+y)`

`(dy)/(dx)=cos^2(x+y)-1`

`(dy)/(dx)=-(1-cos^2(x+y))=-sin^2(x+y)`