# Math Help - Partial Derivative Proof

1. ## Partial Derivative Proof

Hi all,

I have been running through this problem but can't get anywhere near it:

$x = \frac {sin(u)}{cos(v)}$ prove:

$\frac {\partial u}{\partial x} = \frac{cos^3(v)cos(u)}{cos(u+v)cos(u-v)}$

I start off:

$\frac {\partial x}{\partial u} = \frac {cos(u)}{cos(v)}=> \frac {\partial u}{\partial x} = \frac {cos(v)}{cos(u)}$ multiply top and bottom by $cos^2(v)cos(u)$ to give the numerator. Whats left is to prove the denominator:

$cos^2(u) cos^2(v) = cos(u+v)cos(u-v)$

I have tried whatever I can think of, double angle identities,product to sum and even a messy Euler exponential expansion, but fail... is there something fundamental I have missed here? Maybe this should be in the trig sub forum.....

2. I'm glad you couldn't prove $\cos^{2}(u)\cos^{2}(v)=\cos(u+v)\cos(u-v),$ because if you plug in $u=\pi/4=v,$ you don't get equality!

I think your method of differentiation is incorrect. As in, perhaps you're supposed to view $v$ as a function of $x$ as well. What's independent here, and what's dependent?

3. Thanks for looking Ackbeet.... At least I am not going out of my mind!

The wording is suppose $x$ & $y$ are related to $u$ & $v$ via

$x = \frac {sin u}{cos v} ; y = \frac{sin v}{cos u}$

show that (expression in original post) and find a similar expression for $\frac{\partial u}{\partial y}$

I didnt consider that u&v themselves to be functions- that maybe what I needed. I'll give that ago and see what I get.

4. In that case, it's pretty clear that u and v are both functions of x and y. You'll need to use the appropriate formulas for their derivatives.

5. Thanks Ackbeet.

I am rather confused by this problem. Can you by any chance point me in the dirction of the method that will be used so I can look this up..... It looks like a change of variable problem, but it lacks the relation between $x$ and $y$..... do we assume an unknown relation, say $z(x,y)$ define the partials and eliminate the $\partial z$ or is it somrthing different to this?

Any input will be much appreciated...

6. Find u as a function of x and y. Do that by eliminating parameters and using the Pythagorean theorem. I think that path might work for you.

7. This path does work. You can find $\cos^{2}(u)$ in terms of $x$ and $y$, and then use implicit differentiation. You then plug in what $x$ and $y$ are, and simplify to get your final answer. You'll need a trig identity or two, but it does work out.

8. I have had a go at this and get to an identity of:

$cos^2(u) =$ $1- x^2\over{1 - x^2y^2}$

Is this along the lines you took Ackbeet? I guess it would be a case of implicitly differentiating and then plugging original $x$ & $y$ in... is that the route you took? Big thanks to you for helping so far..

9. You should clean up your code a bit. I realize that with the Preview feature taking so long, this isn't very desirable. However, I got the following:

$\displaystyle{\cos^{2}(u)=\frac{1-x^{2}}{1-x^{2}y^{2}}},$

which is, I think, what you're getting at. The route you're suggesting is indeed where I went with it.

10. And we're back. What did you get?