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**implicit differentiation help :)**

## Homework Statement

Find dx/dy by implicit differentiation [itex](x

^{2}+ y)

^{2}+ x

^{2}+ xy

^{2}= 100[/itex]

## Homework Equations

## The Attempt at a Solution

I'm trying to use the chain rule to solve it... i got

The derivative ofb(x

^{2}+ y)

^{2}+ x

^{2}+ xy

^{2}= 100, with respect to x, is 2(x

^{2}+ y)(2x+1*(dy/dx))+ 2x+ y

^{2}+ 2xy(dy/dx)= 0.

This part is right i think (had help getting it) but i'm not sure how to get dy/dx by itself on one side. I'm getting,

2(x

^{2}+ y)(2x+1*(dy/dx))+ 2x+ y

^{2}+ 2xy(dy/dx)= 0

=> dy/dx = -2x-y

^{2}/2(x

^{2}+ y)(2x)(2xy)

I think that is wrong... because when i do the solution by multiplying the squares instead of the chain rule, i get

(x

^{2}+y)

^{2}+x

^{2}+xy

^{2}=100

=(x

^{2}+y)(x

^{2}+y)+x

^{2}+xy

^{2}=100

=(x

^{4}+2x

^{2}y+y

^{2})+x

^{2}+(xy

^{2})=100

4x

^{3}+(2x

^{2}*1)(dy/dx))+y*4x+2y(dy/dx)+2x+(x*2y(dy/dx))+(1*y

^{2})=0

-(4x

^{3}+y*4x+2x+1*y

^{2})=(dy/dx)(2x

^{2}+2y+x*2y)

dy/dx=-(4x

^{3}+y*4x+2x+1*y

^{2})/(2x

^{2}+2y+x*2y)

and if i plug in (x,y) i.e (2,4) i get different answer for the chain rule one.

So what am i doing wrong? I'm pretty sure it's that i am not knowing how to bring the dy/dx on one side the right way using the chain rule

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