How do you differentiate #tanx+tany=1#?

1 Answer
Feb 20, 2017

# dy/dx = - sec^2x/sec^2y #

Explanation:

When we differentiate #y# wrt #x# we get #dy/dx#.

However, we cannot differentiate a non implicit function of #y# wrt #x#. But if we apply the chain rule we can differentiate a function of #y# wrt #y# but we must also multiply the result by #dy/dx#.

When this is done in situ it is known as implicit differentiation.

We have:

# tanx + tany = 1 #

Differentiate wrt #x#:

# sec^2x + sec^2ydy/dx = 0 #
# :. sec^2ydy/dx = - sec^2x #
# :. dy/dx = - sec^2x/sec^2y #