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Quotient & Product Rule: Same funtion, different answers?

  1. Feb 22, 2014 #1
    Can someone check my working. I don't understand why i am getting different answers???
    [tex]u(x,t)=\frac{{e}^{-\frac{x^2}{4Dt}}}{\sqrt{4Dt}}[/tex]
    Differentiate w.r.t 't' by quotient rule:
    [tex]\frac{\partial u}{\partial t}=\left[ \frac{1}{\sqrt{4Dt}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}}-{e}^{-\frac{x^2}{4Dt}}\cdot \frac{1}{\sqrt{4D}}\cdot \frac{-1}{2t\sqrt{t}} \right]\frac{1}{4Dt}[/tex]
    [tex]\frac{\partial u}{\partial t}=\left[ \frac{x^2}{4Dt^2}u(x,t)+\frac{1}{2t}u(x,t)\right]\frac{1}{4Dt}[/tex]
    For the product rule, write as -
    [tex]u(x,t)={\left(4D \right)}^{-\frac{1}{2}}{t}^{-\frac{1}{2}}{e}^{-\frac{x^2}{4Dt}}[/tex]
    Differentiate w.r.t 't' by product rule:
    [tex]\frac{\partial u}{\partial t}=\frac{1}{\sqrt{4D}}\left[ {e}^{-\frac{x^2}{4Dt}}\cdot \frac{-1}{2t\sqrt{t}}+\frac{1}{\sqrt{t}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}} \right][/tex]
    [tex]\frac{\partial u}{\partial t}=\frac{x^2}{4Dt^2}u(x,t)-\frac{1}{2t}u(x,t)[/tex]
    This is clearly not the same.
    What am i doing wrong?
     
    Last edited: Feb 22, 2014
  2. jcsd
  3. Feb 22, 2014 #2

    D H

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    You didn't use the quotient rule correctly. Denoting ##u(x,t)=\frac{v(x,t)}{w(t)}## where ##v(x,t)=\exp\left(\frac {-x^2}{4Dt}\right)## and ##w(t)=\sqrt{4Dt}##, then the quotient rule says
    $$\frac{\partial u(x,t)}{\partial t} = \frac{\frac{\partial v(x,t)}{\partial t}w(t) - v(x,t)\frac{\partial w(t)}{\partial t}}{w(t)^2}$$
     
  4. Feb 22, 2014 #3
    The terms you wrote for [itex]v(x,t)[/itex] and [itex]w(t)[/itex] is what i chose for the quotient rule. The [itex]w(t)^2[/itex] is the [itex]\frac {1}{4Dt}[/itex] term outside the brackets.
    I don't see what am i missing?
     
  5. Feb 22, 2014 #4

    D H

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    You did a couple of things wrong. You computed ##\frac{d}{dt}\sqrt{4Dt}## incorrectly, and you lost a radical term somewhere along the line in your quotient rule computations.
     
  6. Feb 22, 2014 #5
    I've spotted my stupid mistake. Thanks for the pointer. I've been going round in circles with this.
    Cheers
     
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