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- Thread starter paul2001
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HallsofIvy

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The function isHello everyone,

I am trying to take partial derivatives of the following equation and I am having difficulties. The partial derivatives are of Q w.r.t D, ΔP, ρ, and w. Any help would be much appreciated. Thank you.

Paul

[tex]Q= 5.9863 CD^2\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}[/tex]

As a function of D, that is just D

[tex]Q_P= 2(5.9863) CD\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}[/tex]

As a function of [itex]\Delta P[/itex], that is just [itex](\Delta P)^{0.5}[/itex] and the derivative of that is [itex]0.5(\Delta P)^{-0.5}[/itex]

[tex]Q_{\Delta P}= 0.5 (5.9863) CD^2\left[\frac{1}{\Delta P\rho(1-\beta^4)}\right]^{0.5}[/tex]

As a function of [itex]\rho[/itex] it is [itex]1/\rho^{0.5}= \rho^{-0.5}[/itex] and the derivative of that is [itex]-0.5\rho^{-1.5}[/itex]

[tex]Q_{\rho}= -0.5(5.9863) CD^2\left[\frac{\Delta P}{\rho^3(1-\beta^4)}\right]^{0.5}[/tex]

I do not see any "w" in the formula so the derivative with respect to "w" would be 0!

There is a "C" that you did not mention.

[tex]Q_C= 5.9863 D^2\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}[/tex]

There is also a "[itex]\beta[/itex] in the formula. That's a little more complicated because the function involves [itex](1- \beta^4)^{-0.5}[/itex] and the derivative of that, by the chain rule, is [itex]-0.5(1- \beta^4)^{-1.5}(-4\beta^3)= 2(1- \beta^4)^{-1.5}\beta^3[/itex]

[tex]Q_{\beta}= 2(5.9863) CD^2\beta^3\left[\frac{\Delta P}{\rho(1-\beta^{12}}\right]^{0.5}[/tex]

Since this has nothing directly to do with differential equations, I am moving it to "Calculus and Analysis".

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[tex]\sqrt{1 - B^4}[/tex]

using the chain rule since everything else is just a constant multiple that isn't going to be impacted by integration.

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I have attached the same equation with the appropriate substitutions. The problem I am having is that the equation is iterative so I cannot simply take the partial derivatives without solving for Q. Is there a special approach I can use for this type of derivation? Thank you again.

Paul

Paul

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HallsofIvy

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Well, you don't **have** to solve for Q, you can use "implict differentiation".

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Yeah, just put dQ/dwhateva on both sides and then take it like both sides were the expression.

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