Product of function and its derivative

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Discussion Overview

The discussion revolves around the mathematical manipulation of the Darcy equation, specifically the transition from the term involving the product of a function and its derivative to a form involving the derivative of the square of that function. The focus is on understanding the application of the chain rule in this context.

Discussion Character

  • Technical explanation

Main Points Raised

  • Rasmus seeks clarification on the step where h*dh/dx is rewritten as 0.5*d(h^2)/dx.
  • Another participant describes their process of separating variables, integrating, and differentiating to arrive at the expression q = -0.5*K*d(h^2)/dx, questioning the correctness of their approach.
  • A third participant notes that the transformation involves recognizing the chain rule and reversing it, providing the mathematical expression for clarity.
  • Rasmus expresses gratitude for the explanation but questions whether their alternative method, which they find awkward, is still valid.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the validity of Rasmus's alternative method, as it remains unclear whether it is accepted or challenged by others.

Contextual Notes

There may be limitations in understanding the assumptions behind the application of the chain rule and the specific conditions under which the transformations are valid.

RasmusJes
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Hi there

Can someone please explain me this step when rewriting the Darcy equation?

q = -K*h*dh/dx = -0.5*K*d(h^2)/dx

I don't understand how h*dh/dx changes to 0.5*d(h^2)/dxThank you.

-Rasmus
 
Last edited:
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I have tried to separate variables, integrate on both sides, and then differentiate with regard to x:

q = -K*h*dh/dx
<=>
q dx = -K*h*dh
<=>
int q dx = int -K*h*dh
<=>
q*x = -0.5*K*h^2
<=>
d/dx q*x = d/dx (-0.5*K*h^2)
<=>
q = -0.5*K*d(h^2)/dx

Is that correct?
 
It's just recognizing the chain rule and reversing it:

[tex]\frac{dy^2(x)}{dx} = 2y(x)\frac{dy(x)}{dx}[/tex]
 
Wow, that took me a long time to understand :-) Thank you.
Would my suggestion, though much more awkward, still be right?
 

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