How do you differentiate #y=x^2y-y^2-xy#?

2 Answers
Mar 19, 2016

#(dy)/(dx)=(2xy-y)/(1+x-x^2+2y)#

Explanation:

Implicit differentiation is used when a function, here #y#, is not explicitly written in terms #x#. One will need the formula for differentiation of product & ratios of functions as well as chain formula to solve the given function.

For this, we take differential of both sides of the function #y=x^2y-y^2-xy# and differentiating we get

#(dy)/(dx)=d/(dx)(x^2y)-2y(dy)/(dx)-d/(dx)(xy)#

#(dy)/(dx)=2xy+x^2(dy)/(dx)-2y(dy)/(dx)-y-x(dy)/(dx)#

Transposing terms containing #(dy)/(dx)# to LHS we get

#(dy)/(dx)-x^2(dy)/(dx)+2y(dy)/(dx)+x(dy)/(dx)=2xy-y# or

#(dy)/(dx)(1-x^2+2y+x)=2xy-y# or

#(dy)/(dx)=(2xy-y)/(1+x-x^2+2y)#

Mar 19, 2016

Simplify then differentiate to find:

#(dy)/(dx) = 2x-1# when #y != 0#

Multiply by #y/(x^2-x-1)# to cover the case #y=0# and find:

#(dy)/(dx) = ((2x-1)y)/(x^2-x-1)#

Explanation:

graph{y=x^2y-y^2-xy [-10, 10, -5, 5]}

Notice that all of the terms are divisible by #y#, so do that first to find:

#1 = x^2-y-x#

with exclusion #y != 0#

We can rearrange this as:

#y = x^2-x-1#

Hence:

#(dy)/(dx) = 2x-1#

with exclusion #y=0#

What happens in the case #y=0#?

The original equation is satisfied, so its graph consists of the parabola #y = x^2-x-1# together with the x-axis #y=0# for which the derivative is #0#.

So to cover the case #y=0# we can multiply #2x-1# by #y/(x^2-x-1)#, since this has value #1# for all points on the curve where #y != 0# and value #0# when #y=0# (and #x^2-x-1 != 0#).

So: #(dy)/(dx) = ((2x-1)y)/(x^2-x-1)#