Answer
$\frac{{\partial f}}{{\partial r}} = 6rt + 3st - 6{t^2}$
$\frac{{\partial f}}{{\partial t}} = - 12rt + 3{r^2} + 3rs$
Work Step by Step
We are given
$f\left( {x,y,z} \right) = xy + {z^2}$, ${\ \ }$ $x = r + s - 2t$, ${\ \ }$ $y = 3rt$, ${\ \ }$ $z = {s^2}$.
The partial derivatives are
$\frac{{\partial f}}{{\partial x}} = y$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = x$, ${\ \ }$ $\frac{{\partial f}}{{\partial z}} = 2z$
$\frac{{\partial x}}{{\partial r}} = 1$, ${\ \ }$ $\frac{{\partial y}}{{\partial r}} = 3t$, ${\ \ }$ $\frac{{\partial z}}{{\partial r}} = 0$
$\frac{{\partial x}}{{\partial t}} = - 2$, ${\ \ }$ $\frac{{\partial y}}{{\partial t}} = 3r$, ${\ \ }$ $\frac{{\partial z}}{{\partial t}} = 0$
Using the Chain Rule, we calculate the partial derivatives:
1. $\frac{{\partial f}}{{\partial r}}$
$\frac{{\partial f}}{{\partial r}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial r}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial r}} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial r}}$
$\frac{{\partial f}}{{\partial r}} = y + 3xt$
In terms of the independent variables, we get
$\frac{{\partial f}}{{\partial r}} = 3rt + 3\left( {r + s - 2t} \right)t = 6rt + 3st - 6{t^2}$
2. $\frac{{\partial f}}{{\partial t}}$
$\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial t}} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial t}}$
$\frac{{\partial f}}{{\partial t}} = - 2y + 3xr$
In terms of the independent variables, we get
$\frac{{\partial f}}{{\partial t}} = - 6rt + 3\left( {r + s - 2t} \right)r = - 12rt + 3{r^2} + 3rs$
