# Partial diff on inverse tan

1. Jul 28, 2012

### charmedbeauty

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

find dz/dy(partial)

z= tan-1(y/x)

2. Relevant equations

3. The attempt at a solution

z= tan-1(y/x)

let u=y/x

z= tan-1(u)

dz/du = 1/ (1+u2)

so dz/dy = dz/du (du/dy)

du/dy = 1/x

so

dz/dy = (1/1+u2)(1/x)

= 1/ 1+ (y2/x2) * 1/x

= 1/ x + (y2/x2)x)

= x / x + y2

but I should be getting...

dz/dy = x / x2 + y2

but I found dz/dx the same way and it was the right answer.?

whoops I made a arithmetic mistake it is correct.

sorry.

Last edited: Jul 28, 2012
2. Jul 28, 2012

### chiro

Hey charmedbeauty.

Going from the appropriate position I get:

dz/dy = (1/1+u2)(1/x)

= 1/ (1+ (y2/x2)) * 1/x

= 1/ (x + (y^2/x^2)x)

= 1 / (x + y^2/x)

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

= x / [x^2 + y^2]

3. Jul 28, 2012

### szynkasz

Your calculation is ok except the last line

4. Jul 28, 2012

### charmedbeauty

yep for some reason I thought 1/x+(y2/x) = x/ x+y2 but really it did = x/ x2+y2

Thanks.

5. Jul 28, 2012

### HallsofIvy

Staff Emeritus
You did not use the distributive law in the denominator: $(x+ y^2/x)x= x^2+ y^2$

6. Jul 28, 2012

### charmedbeauty

yeah I understand.