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Mathematics 52 Online
OpenStudy (richyw):

If \(z=f(u,v)\) is a continuous function with continuous first and second order derivatives, \(u=xy\) and \(v=\frac{x}{y}\) , use the multivariable chain rule to show that \[x^2\frac{\partial^2z}{\partial x^2}-y^2\frac{\partial^2z}{\partial y^2}=4uv\frac{\partial^2z}{\partial u \partial v}-2v\frac{\partial z}{\partial v}\]

OpenStudy (anonymous):

this is a lot of work

OpenStudy (anonymous):

\[\partial^2z/\partial x^2=?\]

OpenStudy (fwizbang):

Maybe switching variables to logs before you start will make it easier, since ln u = ln x + ln y and ln v = lnx - ln y but mostly you just have to keep applying the chain rule and pushing through the algebra.

OpenStudy (anonymous):

\[\partial z/\partial x= \partial z/\partial u *\partial u/\partial x + \partial z/\partial v *\partial v/\partial x \]

OpenStudy (fwizbang):

\[\partial z/\partial u = g(u,v), etc.....\] so to take the second derivative wrt x you deal with derivatives of derivatives by applying the chain rule again.

OpenStudy (fwizbang):

the good news is that \[{\partial^2 u\over \partial x^2} ={\partial^2 v\over \partial x^2}={\partial^2 u\over \partial y^2}=0\]

OpenStudy (richyw):

what is the question even saying though. is it saying the first order derivative is u=xy and the second order derivative is v=x/y ? because that is what the grammar seems to imply but that doesn't make sense to me

OpenStudy (anonymous):

that doesn't sound right

OpenStudy (anonymous):

z is function composed of u and v; u and v are function composed of x and y

OpenStudy (fwizbang):

The question says that you can think of z either as a function of two independent variables x and y, or that you can think of it as a function of the variables u and v which are related to the first two by the equations given..... i.e. let f(u,v) = u^2 v, then its also true that f(u,v) = x^3y.

OpenStudy (anonymous):

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