# Implicit differentiation problem.

1. Oct 20, 2009

### Dissonance in E

1. The problem statement, all variables and given/known data
Find y´(x) for x + y = 1/x^2 + 8/y^2

2. Relevant equations

3. The attempt at a solution
Rewrite the eq. as
x + y = x^-2 + 8y^-2
differentiate.
1 + y´(x) = (-2x^-3) + (-16y^-3)(y´(x))
Rearrange
1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x))
(1 + (2x^-3))/(16y^-3) = (-2y´(x))
(1 + (2x^-3))/((-2)16y^-3) = y´(x)

Now I know that the answer should look like this:
(1+2x^−3 )/(−1−16y^−3).

What am i doing wrong?

2. Oct 20, 2009

### Fightfish

Something looks really strange for the 2nd line. Careless mistake?

3. Oct 20, 2009

### Dissonance in E

How do i not eliminate the y´(x) term? Im probably doing some really basic mistake that should be enough motivation to make me drop calculus.

1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x))
(1 + (2x^-3))/(-16y^-3) = (y´(x)) - (y´(x))
(1 + (2x^-3))/(-16y^-3) = 0

4. Oct 21, 2009

### Fightfish

You want to make y'(x) the subject, hence you factor it out.
1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x)) = y'(x) (-16y^-3 - 1)
This appears to be an algebra issue that you are facing, so you might want to work on it.

5. Oct 21, 2009

### Tomtom

From: 1 + (2x^-3) = (-16y^-3)(y´(x)) - (y´(x))
You cannot divide by (-16y^-3) on both sides, in order to get your next expression, because the last part of your equation -> " - y´(x) " doesn't have that factor in it.

Instead, "pull out" y´(x) from the second half of the equation giving:

1 + (2x^-3) = y´(x)*[-16y^-3 - 1] (I put in brackets to make it a bit easier.

From here it should be pretty easy, just divide by " -16y^-3 - 1"