How do you find #dy/dx# by implicit differentiation of #x^2-y^3=0# and evaluate at point (1,1)?

1 Answer
Feb 22, 2017

# [dy/dx ]_{ (0,0) } =2/3#

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:

# x^2-y^3=0 #

Differentiate wrt #x#:

# 2x-3y^2dy/dx=0 \ \ \ \ # .... [1]

Although (in this case) we could find an implicit expression for #dy/dx# it is not always necessary, as we are asked to find the value at a particular point.

At #(1,1)# (noting that the point does indeed lie on the initial curve) we have (substituting into [1]):

# 2-3dy/dx=0 => dy/dx=2/3#