Manipulating partial derivitaves/ general derivatives

[AStaunton]

This is a small part in converting between rectangular to polar coords for laplace equation with a problem of circular geometry:

what I have in my notes:

\tan\theta=\frac{y}{x}\implies\sec^{2}\theta\frac{\partial\theta}{\partial x}=\frac{-y}{x^{2}}

I can't figure out how he went from the first part to sec^2theta(del)theta(del)x..

I think an intermediate step is:

\theta=\tan^{-1}\frac{y}{x}

But I do not know how to differentiate the above with respect to x...there is no entry in my log table that corresponds to tan^-1(a/x)...am in a rush to find this out..so any help appreciated.

 

[rock.freak667]

If you differentiate tanθ w.r.t. x, you will get sec²θ dθ /dx

so if you take the partial derivative in that equation you'll get

sec²θ *∂θ /∂x

 

[AStaunton]

thanks, I am not familiar with that result (although I probably should be)..
would you mind explaining a little bit more how differentiating tan(theta) w.r.t.x leads to sec^2..my ability with calculus ain't too strong!

 

[Char. Limit]

You can prove that the derivative of tan(x) is sec²(x) using the quotient rule. Just rewrite tan(x) as sin(x)/cos(x). I'm sure you know the derivatives for those.

 

[AStaunton]

what's confusing me is that if theta=tanx then dtheta/dx=sec^2x <-------that much I know...

but this question is tantheta=1/x...so I don't quite so how it works out the same...again, I stress I am **** at calculus!

 

[Char. Limit]

Well, this is a bit of an abuse of notation, but if we let z=tan(theta), and tan(theta)=y/x, then we can do this:

\frac{dz}{dx} = \frac{dz}{d\theta} \frac{d\theta}{dx}

And since dz/dθ is easy to compute, and so is the derivative on the right side of the equation, we can easily arrive at:

\frac{dz}{d\theta} = sec^2(\theta)

and thus

\frac{dz}{dx} = sec^2(\theta) \frac{d\theta}{dx} = -\frac{y}{x^2}

All you need is the chain rule.

 

[AStaunton]

a follow up question:

in my notes I have:

\cos\theta\frac{\partial}{\partial r}(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta})-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta})=\cos^{2}\theta\frac{\partial^{2}u}{\partial r^{2}}-\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}u}{\partial r\partial\theta}+\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial u}{\partial\theta}-\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}u}{\partial\theta\partial r}+\frac{\sin^{2}\theta}{r}\frac{\partial u}{\partial r}+\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial u}{\partial\theta}+\frac{\sin^{2}\theta}{r^{2}}\frac{\partial^{2}u}{\partial^{2}\theta}

I cannot figure out where the two partial(u)/partial(theta) expressions came from and also where did the partial(u)/partial(r) expression come from?
I clearly don't understand the rules of this properly, my thinking was that we get rid of the brackets by multiplying everything out, but that does not account for the expressions I just mentioned..

 

[AStaunton]

just to repost the right hand side of the equality again as it didn't come out fully:

=\cos^{2}\theta\frac{\partial^{2}u}{\partial r^{2}}-\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}u}{\partial r\partial\theta}+\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial u}{\partial\theta}-\frac{\sin\theta\cos\theta}{r}\frac{\partial^{2}u}{\partial\theta\partial r}+\frac{\sin^{2}\theta}{r}\frac{\partial u}{\partial r}+\frac{\sin\theta\cos\theta}{r^{2}}\frac{\partial u}{\partial\theta}+\frac{\sin^{2}\theta}{r^{2}}\frac{\partial^{2}u}{\partial^{2}\theta}