Question

How do i take the following partial derivative?

Answer 1

partials are simliar to implicits except that you count the nonworking variables as constants.

Answer 2

f(x,y) = xy+y

df/dx = y

df/dy = x+1

Answer 3

\[d(\psi(x,y))/dx = K/(2\pi) \arctan(\arctan(y/x)) \] as well as the partial with respect to y

Answer 4

ouch, thats a mess lol

Answer 5

thanks amistre but i wanted to submit the typed equation.

Answer 6

is arctan tan^-1?

Answer 7

google says YES! haha <3 google

Answer 8

\[f(x,y) = \frac{K}{2\pi\ tan^{-1}(\tan^{-1}(y/x)}\]
i take it K is a constant ?

Answer 9

so how do i handle \[df(x,y)/dx = \tan ^{-1}(\tan^{-1}(y/x))\]

Answer 10

yes K/2pi is going to be a constant

Answer 11

chain rules ...

Answer 12

can we determine what the derivative of: \(\cfrac{1}{tan^{-1}(x)}\) actually is?

Answer 13

u mean 1/tan(x)

Answer 14

thats 1/(1+x^2)

Answer 15

wait hteres a negative

Answer 16

1/ tan-1(x) = y

(1/x) = tan(y)

int: ln|x| = sec^2(y) y'

\(\cfrac{ln|x|}{sec^2(y)} = y'\)

\(\cfrac{ln|x|}{tan^2(1/x)+1} = y'\)

or did i miss something?

Answer 17

man its been along time since calculus 2 for me. lol i cant remember. Um so does each x then equal tan-1(y/x)?

Answer 18

http://www.wolframalpha.com/input/?i=d%28tan%5E-1%28x%29%29%2Fd%28x%29&a=*C.d-_*DerivativesWord-

Answer 19

lol ... when we determine the derivative for 1/tan-1(x) we can address it; but its the chain rule that applys and pops out successive terms on us till we get inside there to deal with the y/x

Answer 20

says d(tan^-1(x))/d(x) can be 1/(x^2+1)

Answer 21

yes, but that is not what you have posted.

http://www.wolframalpha.com/input/?i=d%2Fd%28x%29+%281%2Ftan^-1%28x%29%29

Answer 22

i didnt say 1/tan^-1(x) i said tan^-1(tan^-1(y/x))? or maybe i'm missing a step?

Answer 23

ive gotta run to class; good luck with it :)

Answer 24

dang okay. good luck at class.

Answer 25

I was under the impression that you wanted something close to this:
\[f(x,y) = \frac{K}{2\pi\ tan^{-1}(\tan^{-1}(y/x)}\]
something like this would be easier to do i admit :)
\[f(x,y) = 2\pi\ tan^{-1}(\tan^{-1}(y/x)\]

Answer 26

Ah nope. thanks though. I found a way to do the problem in cylindrical coordinates.

Answer 27

cool :)