Partial Derivative Proof

**padawan** · August 16th 2010, 01:02 PM

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.....

Thanks for reading!

**Ackbeet** · August 16th 2010, 01:25 PM

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?

**padawan** · August 16th 2010, 01:41 PM

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.

**Ackbeet** · August 16th 2010, 03:26 PM

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.

**padawan** · August 17th 2010, 01:00 PM

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...

**Ackbeet** · August 17th 2010, 01:32 PM

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.

**Ackbeet** · August 18th 2010, 08:14 AM

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.

**padawan** · August 18th 2010, 01:21 PM

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..

**Ackbeet** · August 18th 2010, 01:35 PM

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.

**Ackbeet** · August 19th 2010, 04:32 PM

And we're back. What did you get?

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