# Derivative of natural log

1. Oct 16, 2007

### fk378

1. The problem statement, all variables and given/known data
Find y'

y=ln(x^2 + y^2)

2. Relevant equations
d/dx ln(u)= 1/u du/dx

3. The attempt at a solution

y' = [1/(x^2 + y^2)] (2x + 2y)
y' = (2x+2y)/(x^2 + y^2)

But my book says the answer is 2x/(x^2 + y^2 - 2y)

How can that be?

2. Oct 16, 2007

### EnumaElish

You need to be differentiating the y on the right hand side of the equation when you take the derivative of the entire equation.

3. Oct 16, 2007

### fk378

Are you saying my first step should be:
dy/dx = (1/x^2 + y^2) [2x+2y(dy/dx)]

4. Oct 16, 2007

### rocomath

yes

this problem must be solved by implicit differentiation

5. Oct 16, 2007

### fk378

I don't understand what the 2nd step would be though. So I have dy/dx on both sides, but isn't dy/dx what I want to solve for? Wouldn't they cancel each other out?

6. Oct 17, 2007

### HallsofIvy

Staff Emeritus
No, they won't cancel because one is multiplied by $[2y/(x^2+ y^2)]$
(Please, don't write $1/x^2+ y^2$! That's a completely different value).

Solve your equation for dy/dx.

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