Partial derivatives!

but that was not the thing to be proven; it was

x u_x + y u_y = 1, or:

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

yer sorry I forgot to multiply my answers with the x and y

But anyway, I'm not sure how to approach the proof....can anyone offer any help? Is it just the application of log rules?

Well you shouldn't know how to approach the proof, because it is not generally true. Dividing my last equation by x^2+y^2 gets you

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

which is certainly not generally true. In the part of the excercise that that you didn't scan, does it say anything about particular values of x and y?

Oh, but wait a minute! Your derivatives u_x and u_y are wrong. Can you spot the mistakes? (Hint: The derivative of ln(x) is what?)

oh no...they are wrong?

hmm I dunno I can't seem to spot the mistakes...did I use the wrong method of differentiation? I just applied the chain rule by letting:

1/2 ln (u) where u = x^2 + y^2

Applying the chain rule was correct. The mistake was that you used the wrong derivative of the ln function, which is ... what?

oh oh!! omg i can't believe I derived ln (u) wrong

the correct answer should be:

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

u (y) = y/(x^2+y^2)?

oh oh!! omg i can't believe I derived ln (u) wrong

the correct answer should be:

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

u (y) = y/(x^2+y^2)?

And if you multiply x to u(x) and y to u(y), you get:

x^2/(x^2+y^2) and y^2/(x^2+y^2)

add then add the two together, you get:

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

is that right?

That's right. Now you've got it.

ahhh Thanks Thomas! If you hadn't pointed out my silly mistake I would have been at the question for ages

Can I just ask one more question?

If f(x,y,z) = xy^2 - yz^2

One of the questions was to verify that f_x_z = f_z_x (first and second partial derivatives)

I thought the answers were really simple so I'm kind of in doubt whether I've been doing it right or not. For the above question, the answer I got for both f_x_z and f_z_x was zero.

And im not sure if that's right

That's because f is a low degree polynomial (in 3 variables). What you're being asked to verify is "Clairaut's Theorem." It says the mixed 2nd derivatives will be equal if f is well behaved.