Okay your second picture gives the "chain rule":
If f is a function of x, y, and z and x, y, and z are functions of the variables, s and t, then f is a function of s and t and $\frac{\partial f}{\partial t}= f_x\frac{\partial x}{\partial t}+ f_y\frac{\partial y}{\partial t}+ f_z\frac{\partial z}{\partial t}$ and $\frac{\partial f}{\partial s}= f_x\frac{\partial x}{\partial s}+ f_y\frac{\partial y}{\partial s}+ f_z\frac{\partial z}{\partial s}$.
Great!
Now, in the first problem, $f(x, y, z)= xyz^{10}$, $x= t^3$, $y= ln(s^2\sqrt{t})$, and $z= e^{st}$.
The first thing I would do is write $y= 2(1/2)ln(st)= ln(st)$.
Now $f_x= yz^{10}$ and $\frac{\partial x}{\partial t}= 3t^2$ so $f_x\frac{\partial x}{\partial t}= 3t^2yz^{10}$. If you want to reduce that to s and t only, replace y with $ln(st)$ and z with $e^{st}$ to get $f_x\frac{\partial x}{\partial t}= 3t^3ln(sy)e^{10st}$.
$f_y= xz^{10}$ and $\frac{\partial y}{\partial t}= \frac{s}{t}$1 so $f_y\frac{\partial y}{\partial t}= xz^{10}\frac{s}{t}=\frac{st^3e^{10st}}{t}= st^2e^{10t}$.
$f_z= 10xyz^9$ and $\frac{\partial z}{\partial t}=se^{st}$ so $f_z\frac{\partial z}{\partial x}= 10xyz^9(se^{st})= 10(t^3)ln(st)e^{9st}(se^{st})= 90st^3ln(st)e^{10st}$.
$\frac{\partial f}{\partial t}$ is the sum of those.
$\frac{\partial f}{\partial s}$ is done the same way but with $\frac{\partial x}{\partial s}$, $\frac{\partial y}{\partial s}$, and $\frac{\partial z}{\partial s}$.
#3
skeeter
1,103
1
Country Boy said:
The first thing I would do is write $y=2(1/2)ln(st)=ln(st)$
