How to Treat Y When Differentiating Implicitly?

[1MileCrash]

Homework Statement

x^{3}y + y^{3}x = 30

Homework Equations

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The Attempt at a Solution

My understanding is that I'm supposed to take the derivative of each side and try to solve for y'.

But, I don't know how to treat the y when I am differentiating it, to a power or whatever.

For example, here I did

Dx(x^{3}y) + Dx(y^{3}x) = Dx(30)

When differentiating the first term, using the product rule, I have to get Dx(y). Simply saying that it's 1 doesn't seem to work, so I just leave it as y'. Using the product rule on that first term on the left side gives me:

y' + 3(x^2)y.

Added to the derivative of (y^3)(x). But that's the derivative I'm having trouble with.

Can anyone walk me through this? How do I treat y when differentiating a term with it in it? I don't understand.

 

[lanedance]

think of y as a function of x, y = y(x)

then think of the power as a function of y, g(y) = y^3

then you have
y^3 = (y(x))^3 = g(y(x))

now use the chain rule
\frac{d}{dx}y^3 = \frac{d}{dx}g(y(x)) = g'(y(x))y'(x)

 

[1MileCrash]

Will simply deriving it normally and then multiplying the result by y prime always give the same result as the chain rule?

 

[1MileCrash]

Alright, you know that feeling after something very simple clicks, and you kick yourself for not grasping it right off the bat, that's what I'm feeling now. I guess my textbook and I are on two different wavelengths...

x^{3}y+y^{3}x=30

First term differentiated: 3yx^2+y'x^3
Second term: 3xy'y^2 + y^3
derivative of 30 is zero..

So, 3yx^2 + y'x^3 + 3xy'y^2 + y^3 = 0

Since I'm solving for y', get them all to once side..

3yx^2 + y^3 = -y'x^3 - 3xy'y^2

Factor out y' on right side.. : y'(-x^3 - 3xy^2), then divide for a final answer of:

(3yx^2+y^3) / (-x^3 - 3xy^2) = y'

Can someone confirm I'm on the right track?

 

[lanedance]

look good to me
y' = -\frac{y}{x} \frac{(3x^2+y^2} {x^2 + 3y^2}