Question

Find the first partial derratives of u = x/(x^2+y^2)
How do I start, HELP!

Answer 1

So, for instance, when taking partials with respect to x, you treat the other variables variables as constant and proceed as though it were a normal derivative.

In this case, you may want to use the quotient rule, and if it helps, rename y into c to remind you that it is a constant, and then rename the c back to a y after you finish taking the derivative with respect to x.

Answer 2

Likewise when taking the partial with respect to y... i.e., think of x as c.

Answer 3

ah okey, let me try that.

Answer 4

for dx i got
\[\frac{ 2x^2 }{ (x^2+y^2)^2 }\]
which is wrong since by book shows the correct answer to
\[\frac{ -2xy }{ (x^2+y^2)^2 }\]

Answer 5

That doesn't look correct. The numerator should be \[y^2-x^2\].

Answer 6

oh sry I wrote dy not dx, but yes u are correct. and the denominatior is the same for both.
but somehow I went wrong in my calculations, can u make the equation step by step so I can see where I went wrong?
but only for dx.

Answer 7

Let g(x,y) = x
Let h(x,y) = x^2+y^2

Then, by the quotient rule,

\[\frac { \delta } { \delta x } \frac { x } { x^2 + y^2 } = \frac { \delta g(x,y) h(x,y) + g(x,y) \delta h(x,y) } { (h(x,y))^2 }\ =\frac { x^2+y^2 - 2x^2 } { (x^2 + y^2)^2 }\]

Sorry it took so long, first time I really tried using the math latex grammar for input.

Answer 8

The numerator simplifies to y^2 - x^2

Answer 9

Err, with my deltas, I forgot to include the delta x, i.e., \[\frac { \delta g(x,y) } { \delta x }\]

Answer 10

wow that makes perfect sense, how was that formula clear to you?
I can't even find it in my book, thanx man, I'll try with dy now:)

Answer 11

Also, note that \[\frac { \delta g(x,y) } { \delta x } = 1\] and \[\frac { \delta h(x,y) } { \delta x } = 2x\]

Answer 12

Glad I could help!

Answer 13

wait -2x^3 isnt in the answer.

Answer 14

No, is that what you got in the numerator? It should be \[y^2-x^2\]

Answer 15

what u wrote is correct, erase -2x^3 from the numerator.

Answer 16

yes so what happends to -2x^3?

Answer 17

Let's go through this step by step. -2x^3 should never appear in the solution.

Answer 18

Let let g(x,y)=x and h(x,y)=x^2+y^2

Answer 19

And now let's solve the partials for each one of these separately.

Answer 20

Can you find \[\frac { \delta g(x,y) } { \delta x }\] for me?

Answer 21

x/x= 1
but sh/sx is not 2x its 1 because sh is x^2+y^2 and so is sx

Answer 22

so the formula can be shortened to h+g/h(x,y)^2 right?

Answer 23

What's the normal derivative of x^2?

Answer 24

2x

Answer 25

As I mentioned earlier, a partial derivative with respect to x would be like a normal derivative with respect to x, but all the other variations--like y--would be considered to be a constant (instead of doing implicit differentiation).

Answer 26

Thus, \[\frac { \delta } { \delta x } { x^2+y^2} = 2x\]

Answer 27

err, all the other variations == all the other variables

Answer 28

oh yes i got you now. but can u explain the part where -2x^3 dissapears?

Answer 29

Can you show me how you get -2x^3? I never did in my work.

Answer 30

no if we go back to ur equation, the long one, ur next post after that says -2x^3 is then removed so the remaining part for numerator isx^2+y^2, yeS?

Answer 31

I had an x^2+y^2-2x^2, not -2x^3 :)

Answer 32

So, that just equals y^2-x^2

Answer 33

oh sorry yes now that is correct, but I dont just understand why -x^2 is removed?

Answer 34

ohh Now I see, stupid me. I got it man.

Answer 35

Ok. Looks like you just misread the -2x^2 as -2x^3 and that got you on the wrong track for a bit.

Answer 36

-2x^2+x^2 = .-x^2
I never saw that, rookie mistake.

Answer 37

Aye, happens to the best of us. :)

Answer 38

and now for dy, I just change g and h with each other?

Answer 39

No, g and h remain the same, but you take the partials of them wtih respect to y.

So, what's the partial of g(x) = x with respect to y?

Answer 40

0 oh okey I got you.

Answer 41

Remember h(x) and g(x) was just a way to express the function in terms of two functions, h and g, that can be easily matched to the quotient rule.

Answer 42

You could also have just used the product rule, and expressed u as u=x(x^2+y^2)^-1

Answer 43

thats how I did from the beginning but got it all wrong.

Answer 44

That's actually how I did it also in the beginning. I tend to forget the quotient rule :)

Answer 45

Im gonna try fo dy, can u wait? :)

Answer 46

Yup.

Answer 47

I'll be checking in every couple minutes so don't worry if I take a bit to respond.

Answer 48

I solved it!:)

Answer 49

Good job. Starting to click for you now?

Answer 50

yes, thanx to you. Now its time for lunch at 2 pm:)

Answer 51

Bye

Answer 52

I'll probably be back later, so closing this thread now, and thanx again:)