1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taking the derivative

  1. Mar 8, 2010 #1
    In the text (attached) I can't figure out how they are making the jump from the first eqn to the second eqn. Any guidance would be helpful. Thanks
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Apparently, p=p2 and ρ=ρ2, and p is a function of 1/ρ. The quantities q, p1, and ρ1 are constants.

    [tex]\frac{\gamma}{\gamma-1}\left(\frac{p}{\rho}-\frac{p_1}{\rho_1}\right)-\frac{1}{2}\left(\frac{1}{\rho_1}+\frac{1}{\rho}\right)(p-p_1)=q[/tex]

    If you let x=1/ρ, you can write the equation as

    [tex]\frac{\gamma}{\gamma-1}\left(xp(x)-\frac{p_1}{\rho_1}\right)-\frac{1}{2}\left(\frac{1}{\rho_1}+x\right)(p(x)-p_1)=q[/tex]

    Differentiate that equation with respect to x and solve for p'(x).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taking the derivative
  1. Taking derivative (Replies: 2)

  2. Taking the derivative (Replies: 2)

  3. Taking a derivative? (Replies: 2)

  4. Taking derivatives (Replies: 2)

  5. Taking a derivative (Replies: 2)

Loading...