
#1
Aug1108, 09:55 AM

P: 2

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 



#2
Aug1108, 11:50 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

[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^{2} and the derivitive of that is 2D. [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". 



#3
Aug1108, 12:37 PM

P: 75

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:
[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. 



#4
Aug1108, 01:10 PM

P: 2

Need help taking partial derivatives
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 



#5
Aug1108, 01:24 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

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




#6
Aug1108, 01:47 PM

P: 75

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



Register to reply 
Related Discussions  
Second partial derivatives  Calculus & Beyond Homework  9  
taking derivatives?  Programming & Computer Science  4  
MATLAB taking derivatives and plotting  Math & Science Software  0  
Concerning partial derivatives  Calculus  3  
partial derivatives  Calculus  7 