Let $z$ be a function of two variables $u$ and $v$ which are also functions of $x$ and $y$ such that $u=x^{2}y^{3}$ and $v=sin(\pi x)$ I have to compute $ \frac {\partial ^ 2z} { \partial y \partial x } @ (x,y)=(-2,1)$ I am given $z(4,0)=10,$ $z_u(4,0)=5,$ $z_v(4,0)=7,$ $z_{uu}(4,0)=-2,$ $z_{uv}(4,0)=-1,$ $z_{vv}(4,0)=3$ Here is my attempt:
$\frac { \partial z } { \partial x } = \frac { \partial z } { \partial u } \frac { \partial u } { \partial x } + \frac { \partial z } { \partial v } \frac { \partial v } { \partial x }$
Taking the partial derivative of both sides with respect to $y$
$\frac \partial { \partial y }\frac { \partial z } { \partial x } =\frac \partial { \partial y }( \frac { \partial z } { \partial u } \frac { \partial u } { \partial x } + \frac { \partial z } { \partial v } \frac { \partial v } { \partial x })$
which produces
$\frac { \mathrm d ^ 2 z} { \partial y \partial x } = \frac { \partial ^ 2 z } { \partial y \partial u } \frac { \partial u } { \partial x } + \frac { \partial ^ 2 u } { \partial y \partial x } \frac { \partial z } { \partial u } + \frac { \partial ^ 2 z } { \partial y \partial v } \frac { \partial v } { \partial x }+\frac { \partial ^ 2 v } { \partial y \partial x } \frac { \partial z } { \partial v }$
Now, $\frac { \partial u } { \partial x }, \frac { \partial ^ 2 u } { \partial y \partial x }, \frac { \partial z } { \partial u },\frac { \partial v } { \partial x }, \frac { \partial ^ 2 v } { \partial y \partial x }$ are all given or they can be calculated. However $\frac { \partial ^ 2 z } { \partial y \partial u }$ and $\frac { \partial ^ 2 z } { \partial y \partial v }$ are a little different.
$\frac { \partial ^ 2 z } { \partial y \partial u }=\frac \partial { \partial y }\frac { \partial z } { \partial u }=\frac \partial { \partial u }\frac { \partial z } { \partial y } $
$=\frac \partial { \partial u } ( \frac { \partial z } { \partial u } \frac { \partial u } { \partial y } + \frac { \partial z } { \partial v } \frac { \partial v } { \partial y })$
$=\frac \partial { \partial u } ( 5*3xy^{2})$
The calculation of $\frac { \partial ^ 2 z } { \partial y \partial v }$ is similar. What am I gonna do after that point?
This is invalid. You can only swap the order of integration when dealing with independent variables (ie: in the same ordinate system).
Rather you need to use the chain rule again.$$\dfrac{\partial ~~}{\partial y}\dfrac{\partial z}{\partial u}= \dfrac{\partial u}{\partial y}\cdotp\dfrac{\partial^2 z}{\partial u~^2}+\dfrac{\partial v}{\partial y}\cdotp\dfrac{\partial^2 z}{\partial v~\partial u}$$