1. Limited time only! Sign up for a free 30min personal 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!

Solving for du/dx with multiple variables/equations

  1. Aug 14, 2010 #1
    1. The problem statement, all variables and given/known data

    See figure.

    2. Relevant equations


    3. The attempt at a solution

    Okay, my plan of attack to get du/dx was to solve t in terms of x in one equation, solve s in terms of x in the other and all sub intio the original equation and that the derivative with respect to x.

    I can solve t in terms of x,

    [tex]t = \frac{4}{x^3+x^2}[/tex]

    However, I'm having trouble doing the same for s. This is about as far as I can get,

    [tex]s = ln\left(s^3x^2 - 1\right)[/tex]

    Does anyone have any suggestions?

    Attached Files:

  2. jcsd
  3. Aug 14, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    You don't need to solve the equations. Use implicit differentiation to find dt/dx and ds/dx.
  4. Aug 14, 2010 #3
    Okay so,

    [tex]t = \frac{4}{x^3+x^2}[/tex]


    [tex]\frac{dt}{dx} = \frac{-12x-8}{x^5 + 2x^4 + x^3}[/tex]


    [tex]\frac{ds}{dx} = \frac{2xs^3}{e^s-3s^2}[/tex]

    I'm not sure if I solved for ds/dx or dt/dx correct, but I tried.

    Anyone see any problems?
  5. Aug 14, 2010 #4
    This dt/dx is wrong. You somehow managed to lose a factor of x in every single term. I'm guessing it was a careless mistake.

    This ds/dx is correct.

    Do you know how to solve for du/dx from here?
    [tex] \frac{du}{dx} = \frac{\partial u}{\partial s}\frac{ds}{dx} + \frac{\partial u}{\partial t}\frac{dt}{dx} [/tex]
  6. Aug 14, 2010 #5
    I actually factored out an x and canceled it... I didnt know I couldn't do that?
  7. Aug 14, 2010 #6
    You can. Sorry, I guess I just wasn't thinking.
  8. Aug 14, 2010 #7
    Okay since,

    [tex] \frac{du}{dx} = \frac{\partial u}{\partial s}\frac{ds}{dx} + \frac{\partial u}{\partial t}\frac{dt}{dx} [/tex]


    [tex]\frac{du}{dx} = e^te^s\frac{2xs^3}{e^s-3s^2} + (e^te^s - sint)(\frac{-12x-8}{x^5+2x^4+x^3})[/tex]
  9. Aug 14, 2010 #8
    Yep, that sounds right. You could easily solve for t in terms of x and e^s in terms of s and x, but I wouldn't bother really unless the problem specifically says to do so.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook