Question

Help with Implicit Differentiation

Answer 1

\[x ^{2}(x-y)^{2}=x^{2}-y^{2}\]

Answer 2

\[x ^{2}[2(x-y)(1-\frac{ dy }{ dx })] + (x-y)^{2}(2x)=2x-2y \frac{ dy }{ dx }\]

Answer 3

I got that far. I dont know how to deal with the term in the bracket.

Answer 4

x^2 (x-y)^2 = x^2 - y^2

x^2 (x-y) (x-y) = (x+y) (x-y)

x^2 (x-y) = x+y

Answer 5

x^3 - x^2y = x + y

Answer 6

Now Try to Differentiate

Answer 7

working...

Answer 8

\[x^{3}-x^{2}y=x+y\]

Answer 9

that doesn't differentiate into the correct answer according to the solutions manual.

Answer 10

x^3 - x^2 y = x + y

3x^2 - 2xy - x^2.dy/dx = 1 + dy/dx

3x^2 - 2xy - 1 = (x^2+1)dy/dx

dy/dx = (3x^2-2xy-1)/(x^2+1)

Answer 11

I know.. whe you factor out the (x-y), you assumed \(x\ne y\) and thus narrowed your domain..
you were on the right track before..
expand the factors and keep \(\Large {dy\over dx}\) terms together and the rest on the other side.

Answer 12

that is not the correct answer. That is what i got when i differentiated what u gave.

Answer 13

but...@electrokid wat is wrong in it ?

Answer 14

If I type the next step in the sol manual.. can u tell me how they got from one step to the next? It seems if I could only deal with the terms in the brackets I can do the rest

Answer 15

@yahoo! you cannot cancel out the terms as you wish. they have certain conditions..
you can cancel out two term iff the term is non-zero.

Answer 16

\[-2x^2(x-y)\frac{ dy }{ dx }+2y \frac{ dy }{ dx }=2x-2x^2(x-y)-2x(x-y)^2\]

Answer 17

\[
2x^2(x-y)-2x^2(x-y){dy\over dx}+2x(x-y)^2=2x-2y{dy\over dx}\\
\cancel{2}x^2(x-y)+\cancel{2}x(x-y)^2-\cancel{2}x=\cancel{2}x^2(x-y){dy\over dx}-\cancel{2}y{dy\over dx}\\
x(x-y)\left[x+(x-y)\right]=\left[x^2(x-y)-y\right]{dy\over dx}\\
\dots
\]

Answer 18

so i just distribute the x^2 to the 2 to both terms inside the brackets?

Answer 19

you may or may not..
divide both sides by
\(x^2(x-y)-y\)

Answer 20

that will give dy/dx right away

Answer 21

If you look way up to my second post, how do I deal with the terms inside the brackets? []

Answer 22

I think I missed a square in the equation!!

Answer 23

since they are all multiplying, the square-parantheses do not have any meaning

Answer 24

so, what I did was expanded to get the "dy/dx" out.. that was my first step. compare my first step to your second post.
I started from your second post

Answer 25

how did u get -2x^2(x-y)?

Answer 26

the second term

Answer 27

thats what im missing