# Implicit Differentiation

## Homework Statement

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

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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.

## Answers and Replies

lanedance
Homework Helper
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
$$\frac{d}{dx}y^3 = \frac{d}{dx}g(y(x)) = g'(y(x))y'(x)$$

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

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