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Math Help - Proof Involving Partial Derivatives Chain Rule

  1. #1
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    Proof Involving Partial Derivatives Chain Rule

    Hi,

    "if u=f(x,y) is a differentiable function with x=e^s((cos(t)) y=e^s((sin(t)) show that show d2u/dx2+d2u/dy2 = e^(-2s)[d2u/ds2+ d2u/dt2

    I know how to find dx/dt dy/ds etc, but I don't know how to find anything involving u because it is only defined as f(x,y).

    Thank your for your help.
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  2. #2
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    Quote Originally Posted by runningeagle View Post
    Hi,

    "if u=f(x,y) is a differentiable function with x=e^s((cos(t)) y=e^s((sin(t)) show that show d2u/dx2+d2u/dy2 = e^(-2s)[d2u/ds2+ d2u/dt2

    I know how to find dx/dt dy/ds etc, but I don't know how to find anything involving u because it is only defined as f(x,y).
    Use the chain rule: \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}. Then when you differentiate a second time you have to use the product rule as well as the chain rule, for example \frac{\partial }{\partial s}\Bigl(\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}\Bigr) = \Bigl(\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y\partial x}\frac{\partial y}{\partial s}\Bigr)\frac{\partial x}{\partial s} + \frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial s^2}. That way, you can find an expression for \frac{\partial^2 u}{\partial s^2} + \frac{\partial^2 u}{\partial t^2}.
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