Implicit Differentiation: Solving for dx/dy in an Example Problem

[A_Munk3y]

Let me first say i just learned implicit differentiation (earlier today) and i am also new in general to derivatives. I am finding implicit differentiation difficult but i want to understand it before we go over it in class.

Homework Statement

This is a example in my book. I have been trying to solve it for a while now, but I'm not sure i am doing it right and the book does not have answers so i can not check if i am even close to the right answer.
Find dx/dy by implicit differentiation (x²+y)²+x²+xy²=100

The Attempt at a Solution

(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)
I don't think that is right is it?

 

[JThompson]

See attached JPEG. If you need more help, just ask.

 

[A_Munk3y]

fixed it up... is that better now?
i removed the equal sign too, but didn't right anything on here (makes it confusing) :)
(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)

 

[HallsofIvy]

Yes, that is correct. You could also do this without multiplying out the squares, using the chain rule:
$$((x^2+ y)^2)'= 2(x^2+ y)(x^2+ y)'= 2(x^2+y)(2x+ y')$$
while the derivative of $x^2+ xy^2$, with respect to x, is $2x+ y^2+ 2xyy'$.

The derivative of $(x^2+ y)^2+ x^2+ xy^2= 100$, with respect to x, is [itex]2(x^2+ y)(2x+y')+ 2x+ y^2+ 2xyy'= 0[/math].

Notice that I say "is given by". To find the actual derivative of y with respect to x, you solve for y' as you did.

 

[A_Munk3y]

Ahh... the chain rule :) I have such a hard time with that rule! lol
But thanks for the info. I don't really understand how you did that, but i will look up the rule again and then come back and see how you went about doing what you did.

Thanks a lot!