Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: What is wrong with this differentiation?

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Find dy/dt, given:
    [tex]y=x^2-5x+1[/tex]
    [tex]x=s^3-2s+1[/tex]
    [tex]s=\sqrt{t^2+1}[/tex]

    2. The attempt at a solution
    I am trying to use the chain rule for the case where y is a function of x, x is a function of s and s is a function of t, like below:
    [tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{ds}\frac{ds}{dt}[/tex]
    So:
    [tex]\frac{dy}{dt}=(2x-5)(3s^2-2)(\frac{t}{\sqrt{t^2+1}})[/tex]
    But this gives (substituting x and s for their values as functions of t)
    [tex]6t^5-4t^3-\frac{3t}{\sqrt{t^2+1}}-\frac{9t^3}{\sqrt{t^2+1}}-2t[/tex]
    which is not the correct result. The correct result is:
    [tex]6t^5-4t^3-9t\sqrt{t^2+1}+\frac{6t}{\sqrt{t^2+1}}-2t[/tex]
    What is wrong?

    Thank you in advance.
     
  2. jcsd
  3. Jul 7, 2011 #2
    Never mind. After doing some manipulations I found that apparently both results are equal, because these two terms of the answer given as correct:
    [tex]-9t\sqrt{t^2+1}+\frac{6t}{\sqrt{t^2+1}}[/tex]
    after some manipulation:
    [tex]-9t\sqrt{t^2+1}+\frac{6t}{\sqrt{t^2+1}}=\frac{-9t(t^2+1)+6t}{\sqrt{t^2+1}}=\frac{-9t^3-3t}{\sqrt{t^2+1}}[/tex]
    become the two terms of the answer I found.
     
    Last edited: Jul 7, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook