Need help taking partial derivatives

[paul2001]

Hello 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

 

[HallsofIvy]

Hello 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

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

As a function of D, that is just D² and the derivitive of that is 2D.
Q_P= 2(5.9863) CD\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}

As a function of \Delta P, that is just (\Delta P)^{0.5} and the derivative of that is 0.5(\Delta P)^{-0.5}
Q_{\Delta P}= 0.5 (5.9863) CD^2\left[\frac{1}{\Delta P\rho(1-\beta^4)}\right]^{0.5}

As a function of \rho it is 1/\rho^{0.5}= \rho^{-0.5} and the derivative of that is -0.5\rho^{-1.5}
Q_{\rho}= -0.5(5.9863) CD^2\left[\frac{\Delta P}{\rho^3(1-\beta^4)}\right]^{0.5}

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

There is also a "\beta in the formula. That's a little more complicated because the function involves (1- \beta^4)^{-0.5} and the derivative of that, by the chain rule, is -0.5(1- \beta^4)^{-1.5}(-4\beta^3)= 2(1- \beta^4)^{-1.5}\beta^3
Q_{\beta}= 2(5.9863) CD^2\beta^3\left[\frac{\Delta P}{\rho(1-\beta^{12}}\right]^{0.5}

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

 

[Alex6200]

Just pretend that whatever variable you're not taking the partial derivative with respect to is a constant. For example if you're taking it with respect to B, just treat all of the other variables like constants, and differentiate it the way you would:

\sqrt{1 - B^4}

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

 

[paul2001]

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

 

[HallsofIvy]

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

 

[Alex6200]

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