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!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook