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Derivatives and differentials

  • Thread starter Niles
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[SOLVED] Derivatives and differentials

1. Homework Statement
Hmm, when I have


[tex]\frac{dx^2}{dx}[/tex], does this equal zero or 2x?

What confuses me is the way it is written.
 

Answers and Replies

1,750
1
Uh ... what is the original problem? And did you copy that exactly?

1st derivative = [tex]\frac{d}{dx}[/tex]

2nd derivative = [tex]\frac{d^2}{dx^2}[/tex]

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

It's not like [tex]\frac{dy}{dx}[/tex] ... which states that you're taking the derivative of y with respects to x.
 
Last edited:
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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?
 
1,750
1
That's beyond me ... :p
 
1,868
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Ok, but thanks for taking the time to look at it.

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

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