Answer
a) $\frac{{\partial f}}{{\partial x}} = 2x{y^3}$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = 3{x^2}{y^2}$, ${\ \ }$ $\frac{{\partial f}}{{\partial z}} = 4{z^3}$
(b) $\frac{{\partial x}}{{\partial s}} = 2s$, ${\ \ }$ $\frac{{\partial y}}{{\partial s}} = {t^2}$, ${\ \ }$ $\frac{{\partial z}}{{\partial s}} = 2st$
(c) $\frac{{\partial f}}{{\partial s}} = 7{s^6}{t^6} + 8{s^7}{t^4}$
Work Step by Step
(a) We have $f\left( {x,y,z} \right) = {x^2}{y^3} + {z^4}$ and $x = {s^2}$, $y = s{t^2}$, and $z = {s^2}t$.
The primary derivatives are
$\frac{{\partial f}}{{\partial x}} = 2x{y^3}$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = 3{x^2}{y^2}$, ${\ \ }$ $\frac{{\partial f}}{{\partial z}} = 4{z^3}$
(b) Calculate
$\frac{{\partial x}}{{\partial s}} = 2s$, ${\ \ }$ $\frac{{\partial y}}{{\partial s}} = {t^2}$, ${\ \ }$ $\frac{{\partial z}}{{\partial s}} = 2st$
(c) Compute $\frac{{\partial f}}{{\partial s}}$ using the Chain Rule:
$\frac{{\partial f}}{{\partial s}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial s}} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial s}}$
Using the results from part (a) and part (b) and substituting them in the equation above we get
$\frac{{\partial f}}{{\partial s}} = 4x{y^3}s + 3{x^2}{y^2}{t^2} + 8{z^3}st$
Since $x = {s^2}$, $y = s{t^2}$, and $z = {s^2}t$, so $\frac{{\partial f}}{{\partial s}}$ in terms of the independent variables $s$ and $t$ is
$\frac{{\partial f}}{{\partial s}} = 4\left( {{s^2}} \right){\left( {s{t^2}} \right)^3}s + 3{\left( {{s^2}} \right)^2}{\left( {s{t^2}} \right)^2}{t^2} + 8{\left( {{s^2}t} \right)^3}st$
$ = 4{s^6}{t^6} + 3{s^6}{t^6} + 8{s^7}{t^4}$
$ = 7{s^6}{t^6} + 8{s^7}{t^4}$
