Question

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 \[x^2\frac{\partial^2z}{\partial x^2}-y^2\frac{\partial^2z}{\partial y^2}=4uv\frac{\partial^2z}{\partial u \partial v}-2v\frac{\partial z}{\partial v}\]

Answer 1

this is a lot of work

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

that doesn't sound right

Answer 9

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

Answer 10

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.