How do you differentiate #sinx-siny=x-y-5#?

Question

1252 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-29T19:05:20

# dy/dx = (cosx - 1)/(cosy - 1)#

# sinx-siny=x-y-5 #

Differentiate wrt #x#
# :. d/dxsinx-d/dxsiny=d/dxx - d/dxy - d/dx5 #
# :. cosx - d/dxsiny=1 - dy/dx #

We can't differentiate #siny# wrt #x# but we can differentiate wet y using the chain rule (this is the implicit differentiation)

# :. cosx - dy/dxd/dysiny=1 - dy/dx #

# :. cosx - dy/dxcosy=1 - dy/dx #

We can now collect terms and factorise:
# :. dy/dxcosy - dy/dx = cosx - 1#

# :. dy/dx(cosy - 1) = cosx - 1#

# :. dy/dx = (cosx - 1)/(cosy - 1)#