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Homework Help: 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


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    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.
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