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

Question

Christopher Allers · Accepted Answer

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

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#

Emilia Aguilar · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation of $ x^2 - y^3 = 0 $, differentiate both sides with respect to $ x $, then solve for $ \frac{dy}{dx} $.

$ \frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(y^3) = 2x - 3y^2 \frac{dy}{dx} $

$ \frac{dy}{dx}(1,1) = 2(1) - 3(1)^2 \frac{dy}{dx} = 2 - 3\frac{dy}{dx} = 0 $

$ \frac{dy}{dx} = \frac{2}{3} $ at point (1,1).