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Multivariable chain rule question

  1. Jan 7, 2010 #1
    I am trying to find the second derivative of the function

    [tex] C:[0,1]^{2} \rightarrow [0,1] ,\quad \mbox{defined by }C=C(u,v) [/tex]

    evaluated at

    [tex]u=F(x)=1-\exp(-\lambda_{1} x),\quad \lambda_{1} \geq 0 [/tex]

    and

    [tex]v=G(x)=1-\exp(-\lambda_{2} x),\quad \lambda_{2} \geq 0[/tex]

    First I work out the first derivative which is

    [tex]\dfrac{dC}{dx} = \dfrac{\partial C}{\partial u}\dfrac{du}{dx}+\dfrac{\partial C}{\partial v}\dfrac{dv}{dx}[/tex]

    Now, I have trouble working out the second derivative because it looks like I have to used the chain rule again and there is product rule which involves differentiating

    [tex]\dfrac{du}{dx}[/tex]

    with respect to u(and v)??


    I would appreciate any reply. Thank you guys.
     
    Last edited: Jan 7, 2010
  2. jcsd
  3. Jan 7, 2010 #2

    Landau

    User Avatar
    Science Advisor

    Please put tex-tags around your latex expressions.

    [tex]\dfrac{dC}{dx} = \dfrac{\partial C}{\partial u}u'+\dfrac{\partial C}{\partial v}v'[/tex]

    First use the product (and sum) rule:

    [tex]\dfrac{d^2C}{dx^2} = \left(\frac{d}{dx}\dfrac{\partial C}{\partial u}\right)u'+\frac{\partial C}{\partial u}u''+\left(\frac{d}{dx}\dfrac{\partial C}{\partial v}\right)v'+\frac{\partial C}{\partial v}v''[/tex]

    Now use the chain rule to write out the epressions

    [tex]\left(\frac{d}{dx}\dfrac{\partial C}{\partial u}\right)[/tex]
    [tex]\left(\frac{d}{dx}\dfrac{\partial C}{\partial v}\right)[/tex]
     
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