Question

find dy/dx by implicit differentiation: x^2[x-y]^2=x^2-y^2

Answer 1

Do you know how to differentiate xy with respect to x?

Answer 2

yes I just confused after doing the chain rule

Answer 3

Would you mind showing what you have got so far?

Answer 4

x^2[x-y]^2=x^2-y^2

2x[x-y]^2+x^2(2)(x-y)(1-dy/dx)=2x-2ydy/dx

2x[x-y]^2+x^2(2)(x-y)(1-dy/dx)+2ydy/dx=2x

1-dy/dx+2ydy/dx=2x-2x[x-y]^2-x^2(2)(x-y)

Answer 5

\[x^2(x-y)^2=x^2-y^2\]
^This is the question, right?

Answer 6

yes

Answer 7

First, let's simplify it a bit.
Consider the right side, can you factorize x^2 - y^2?

Answer 8

the derivative of the right side is 2x-2ydy/dx

Answer 9

Denote dy/dx as y'
\[x^2(x-y)^2=x^2-y^2\]Differentiate both sides w.r.t. x
\[2x[x-y]^2+x^2(2)(x-y)(1-y')=2x-2yy'\]
Expand the term \(2x^2(x-y)(1-y')\)
\((2x^3 - 2x^2y)(1-y') = 2x^3-2x^2y - y'(2x^3 - 2x^2y)\)

So, we get
\[2x[x-y]^2+ 2x^3-2x^2y - y'(2x^3 - 2x^2y)=2x-2yy'\]
Put the terms with y' on the same side:
\[2x[x-y]^2+ 2x^3-2x^2y - 2x= y'(2x^3 - 2x^2y)-2yy'\]
Take out y' as the common factor and isolate it.

Answer 10

Another way I am thinking is:
\[x^2[x-y]^2=x^2-y^2\]Factorize the right side, we get
\[x^2[x-y]^2=(x-y)(x+y)\]
Divide both sides by x-y, for x≠y. Then, we get
\[x^2(x-y)=x+y\]Expand the left side
\[x^3-x^2y=x+y\]Now we can differentiate the equation.

Answer 11

I think the your second method is great idea

Answer 12

a lot faster and for someone who doesn't understand implicit differentiation very well this method allows you to see where it comes into play

Answer 13

Thank you so much

Answer 14

Usually, it is better if you can simplify the expression a bit :)
You're welcome~

Answer 15

can i ask you another math question

Answer 16

Yes :) But better ask a new question in a new post

Answer 17

ok thanks

Answer 18

Welcome again