How do you find #dy/dx# by implicit differentiation of #x^2-y^2=16#?

1 Answer
Nov 23, 2016

# dy/dx = x/y #

Explanation:

If an equation does not express #y# explicitly in terms of #x# as in #y=f(x)#, and instead we have #g(y)=f(x)#, then when we differentiate we apply the chain rule so that we differentiate #g(y)# wrt #y# rather than #x# as in:

# g(y) = f(x) #
# :. d/dx g(y) = d/dx f(x) #
# :. dy/dx d/dy g(y) = f'(x) #
# :. g'(y) dy/dx= f'(x) #

This will typically result in #dy/dx# being a function of #x# and #y#, rather than #dy/dx# being a function of #x# alone as we usually get

So for # x^2 - y^2 = 16 # we have:

# d/dx(x^2) - d/dx(y^2) = d/dx(16) #
# :. 2x - dy/dxd/dy(y^2) = 0 #
# :. 2x - 2ydy/dx = 0 #
# :. 2ydy/dx = 2x #
# :. dy/dx = x/y #