# Need help taking partial derivatives

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

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HallsofIvy
Homework Helper
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".

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.

HallsofIvy