Derivatives and Differentials: Solving for \frac{dx^2}{dx}

[Niles]

[SOLVED] Derivatives and differentials

Homework Statement

Hmm, when I have


\frac{dx^2}{dx}, does this equal zero or 2x?

What confuses me is the way it is written.

 

[rocomath]

Uh ... what is the original problem? And did you copy that exactly?

1st derivative = \frac{d}{dx}

2nd derivative = \frac{d^2}{dx^2}

I think you meant ... \frac{d}{dx}(x^2)=2x (which says ... this is the derivative of x ...) <--- just an example!

It's not like \frac{dy}{dx} ... which states that you're taking the derivative of y with respects to x.

 

[Niles]

The original problem is:

Consider the 2D Laplace equation in polar cylindricals. Assume the solutions u(rho, Phi) = rho^n * Phi(phi), where n > 0.

I have to find u(rho, Phi).

What they do in the solution is to find the solution for Phi(phi) = A*cos(...) + B*sin(...), and then they set the total solution u(rho, Phi) = \sum [ A*cos(...) + B*sin(...) ] * rho^n.

So I got confused. They do not find the solution for rho^n, but they just multiply it on? That doesn't make sense since we have to take the deivate of rho in Laplace's eq. in 2D?

 

[rocomath]

That's beyond me ... :p

 

[Niles]

Ok, but thanks for taking the time to look at it.

If anybody else has a suggestion, I am all ears.