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Implicit differentiation help

  • Thread starter A_Munk3y
  • Start date
  • #1
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implicit differentiation help :)

Homework Statement


Find dx/dy by implicit differentiation [itex](x2+ y)2+ x2+ xy2= 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(x2+ y)2+ x2+ xy2= 100, with respect to x, is 2(x2+ y)(2x+1*(dy/dx))+ 2x+ y2+ 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(x2+ y)(2x+1*(dy/dx))+ 2x+ y2+ 2xy(dy/dx)= 0

=> dy/dx = -2x-y2/2(x2+ 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
(x2+y)2+x2+xy2=100
=(x2+y)(x2+y)+x2+xy2=100
=(x4+2x2y+y2)+x2+(xy2)=100
4x3+(2x2*1)(dy/dx))+y*4x+2y(dy/dx)+2x+(x*2y(dy/dx))+(1*y2)=0
-(4x3+y*4x+2x+1*y2)=(dy/dx)(2x2+2y+x*2y)
dy/dx=-(4x3+y*4x+2x+1*y2)/(2x2+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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Answers and Replies

  • #2
128
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This

2(x2+ y)(2x+1*(dy/dx))+ 2x+ y2+ 2xy(dy/dx)= 0
=> (dy/dx)= -2x-y2/ 2(2x2+y)(2x+1)(2xy)

Looks strange.

To solve it, expand the whole thing collect all terms that have a factor dy/dx on one side the rest on the other. Factorize out dy/dx and divide away the factor in front of it.
 
  • #3
72
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That's what i'm trying to do :)
Except I'm failing at it lol...

I'll show how i went about it step by step so you can correct where i went wrong
2(x2+ y)(2x+1*(dy/dx))+ 2x+ y2+ 2xy(dy/dx)= 0
=> 2(x2+ y)(2x+1*(dy/dx))+2xy(d)y/dx)= -2x-y2
=> 2x+1*(dy/dx)+2xy(dy/dx)= -2x-y2/2(x2+ y)
=> 1*(dy/dx)+2xy(dy/dx)= -2x-y2/2(x2+ y)(2x)
=> dy/dx = -2x-y2/2(x2+ y)(2x)(2xy)
 
  • #4
128
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[tex] 2(x^2+ y)(2x+1(dy/dx)) [/tex]
[tex] = 4x^3 + 2x^2(dy/dx) + 4yx + 2y(dy/dx) [/tex]

Right?

You can't just divide away the [tex] 2(x^2 + y) [/tex] because it is not a common factor on the left hand side.

If you had ab + c = d you can't just say that a + c = d/b.
 
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  • #5
72
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Oh! lol, i didn't know that :) Thanks for the info!

So is this correct?
2(x2+y)(2x+1*(dy/dx))+2x+y2+2xy(dy/dx)=0
=> 4x3+2x2(dy/dx)+2yx+2y(dy/dx)+2x+y2+2xy(dy/dx)=0
=> (dy/dx)*(2x2+2y+2xy)= -4x3-2yx-2x-y2
=> dy/dx = -4x3-2yx-2x-y2/ (2x2+2y+2xy)


because i'm still getting a different answer :(
i got dy/dx=-(4x3+y*4x+2x+1*y2)/(2x2+2y+x*2y) using multiplying the squares.
 
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  • #6
128
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In the second line you forgot the 2xy term but then it popped back on the third line in the right hand side.

The answer in the end looks sort of right.
 
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  • #7
72
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yea, i forgot to put it in, but i knew it was there cause i had it right on my paper.
Question though:

For this im getting --- (dy/dx)= -4x3-2yx-2x-y2/ (2x2+2y+2xy)
but when i multiplied the squares instead of using the chain rule, i got
dy/dx = -(4x3+y*4x+2x+1*y2)/(2x2+2y+x*2y)
dy/dx = -4x3-4xy-2x-y2/ (2x2+2y+2xy)


They are very similar, but they are still different (the difference is the 2xy//4xy in the numerator). Did i do the math wrong somewhere?
 
  • #8
128
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I made a mistake when getting the 2xy term. It should have been 4xy. Then they are the same.

I fixed my stuff above. (You might need to refresh the page to see the changes).
 
  • #9
72
0


Oh :) Ok i see where it is.
Well, thanks a lot man! I really appreciate all the help
 

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