Question

If z = arctan(y/x), find d^2z/dxdy

Answer 1

\[\huge z \ = \ \tan^{-1}\left(\frac{y}x\right)\]
And you're to find
\[\huge \frac{\partial^2z}{\partial x \partial y}\]
right?

Answer 2

Partial derivatives... first, find its derivative with respect to y (or x, really, but I think y is better)
Find its derivative with respect to y whilst pretending for the moment that x is a constant.

Answer 3

Yes that's what I meant, sorry didn't see the "Equation" button, woops.

Answer 4

So... can you find the derivative with respect to y?

Answer 5

\[\frac{1}{ 1 + (\frac{ Y }{ X })^{2}}\] Am I on the right track?

Answer 6

Quite so :)
Now take the derivative of THIS expression, this time with respect to x. And THIS time, pretend y is a constant.

Answer 7

so, in this notation its read from right to left, correct?

dy then dx

Answer 8

Are you referring to the order in which this is differentiated?

Answer 9

I'm having a blank moment :/

Answer 10

At differentiating this function:
\[\huge \frac{\partial}{\partial x}\frac{1}{1+\left(\frac{y}x\right)^2}\]?

Answer 11

Yes

Answer 12

Okay, chain rule... first, the derivative of 1/x, we apply it here...
\[\huge -\frac{1}{\left(1+\left(\frac{y}x\right)^2\right)^2}\cdot\frac{\partial}{\partial x}\left(1+\left(\frac{y}{x}\right)^2\right)\]
Does this make you feel better? :D

Answer 13

\[D_y[tan^{-1}(ky)]=\frac{k}{1+(ky)^2}\]

Answer 14

Much better :D and to get 2nd derivative I just derive again

Answer 15

Yes. The word is differentiate, not derive, by the way :D

Answer 16

Alright woops my bad, terrible text book this

Answer 17

Lol, that's semantics, this is mathematics. Please proceed :)

Answer 18

your derivative is missing the constant (1/x) ... unless ive overlooked it

Answer 19

I think the y/x does that?

Answer 20

no. what is the derivative w.r.t.y of say: arctan(6y) ?

Answer 21

\[D_y[\tan^{-1}(6y)]\ne\frac{1}{1+(6y)^2}\]

Answer 22

Snap... You're RIGHT
@MathsNoob URGENT
We forgot the 1/x bit when we differentiated with respect to y!

Answer 23

it happens :)

Answer 24

This is why I hate calculus... actually, I don't have a reason, I just hate it :D

Answer 25

see if i can write this up as a product rule now ....

\[D_x[x^{-1}~(1+kx^{-2})^{-1}]\]

Answer 26

such that k=y^2

Answer 27

So... \[z = \tan ^{-1}(\frac{ Y }{ X })\]
\[= (\frac{ 1 }{ 1 +(\frac{ Y }{ X })^{2} }). (-\frac{ Y }{ X ^{2} }) + (\frac{ 1 }{ 1 +(\frac{ Y }{ X })^{2} }). (-\frac{ 1 }{ X })\]
???

Answer 28

it might prove better to simplify Zy first

\[\frac{\frac 1x}{1+\frac{y^2}{x^2}}\]
\[\frac{x^2}{x^2}\frac{\frac 1x}{1+\frac{y^2}{x^2}}\]
\[\frac{x}{x^2+y^2}\]

Answer 29

\[D[fg]=f'g+fg'\]
\[f=x~:~f'=1\]
\[g=(x^2+k)^{-1}~:~g'=-(x^2+k)*2x\]

Answer 30

-(x^2+k)^(-2) that is :)

Answer 31

\[Z_{yx}=1(x^2+y^2)^{-1}+x(-2x)(x^2+y^2)^{-2}\]
\[Z_{yx}=[(x^2+y^2)+x(-2x)](x^2+y^2)^{-2}\]
\[Z_{yx}=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}\]