Partial Differentiation of Inverse Tangent Function

[charmedbeauty]

Homework Statement

find dz/dy(partial)

z= tan^-1(y/x)

Homework Equations

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 + y²but 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.

 

[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]

 

[szynkasz]

Your calculation is ok except the last line

 

[charmedbeauty]

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]

Your calculation is ok except the last line

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

Thanks.

 

[HallsofIvy]

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

 

[charmedbeauty]

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

yeah I understand.