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\centerline
{
\bf Dr. Z's Math251 Handout \#14.6 (2nd ed.) [The Chain Rule]
}

By Doron Zeilberger

{\bf Problem Type 14.6a}: 
Use the chain rule to find
${{\partial z} \over {\partial s}}$
and ${{\partial z} \over {\partial t}}$ :
$$
z=f(x,y) \quad , \quad x=h_1(s,t) \quad , \quad y=h_2(s,t) \quad .
$$

{\bf Example Problem 14.6a}: Use the chain rule to find
${{\partial z} \over {\partial s}}$
and ${{\partial z} \over {\partial t}}$ :
$$
z=x^3+2xy+y^2 \quad , \quad x=s+2t \quad , \quad y=s^2 t \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Set-up the chain-rule:
$$
{{\partial z} \over {\partial s}}=
{{\partial z} \over {\partial x}}{{\partial x} \over {\partial s}}
+
{{\partial z} \over {\partial y}}{{\partial y} \over {\partial s}}
\quad ,
$$
$$
{{\partial z} \over {\partial t}}=
{{\partial z} \over {\partial x}}{{\partial x} \over {\partial t}}
+
{{\partial z} \over {\partial y}}{{\partial y} \over {\partial t}}
\quad .
$$

&
{\bf 1.} 
Ditto.
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{\bf 2.} For the specific functions given compute all
the necessary quantities that show up on the right sides,
namely:
${{\partial z} \over {\partial x}}$,
${{\partial z} \over {\partial y}}$,
${{\partial x} \over {\partial s}}$,
${{\partial y} \over {\partial s}}$,
${{\partial x} \over {\partial t}}$,
${{\partial y} \over {\partial t}}$.

&
{\bf 2.} 
$$
{{\partial z} \over {\partial x}}=3x^2+2y \quad , \quad
{{\partial z} \over {\partial y}}=2x+2y \quad .
$$
$$
{{\partial x} \over {\partial s}}=1 \quad, \quad
{{\partial y} \over {\partial s}}=2st \quad .
$$
$$
{{\partial x} \over {\partial t}}=2 \quad , \quad
{{\partial y} \over {\partial t}}=s^2  \quad .
$$
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{\bf 3.} Incorporate them into the fomulas of step 1.

&
{\bf 3.} 
$$
{{\partial z} \over {\partial s}}=
(3x^2+2y) \cdot 1
+(2x+2y)(2st)=3x^2+2y+4xst+4yst \quad.
$$
$$
{{\partial z} \over {\partial t}}=
(3x^2+2y) \cdot 2+(2x+2y)s^2=
6x^2+4y+2xs^2+2ys^2 \quad.
$$
{\bf Ans.:}
${{\partial z} \over {\partial s}}=3x^2+2y+4xst+4yst$,
${{\partial z} \over {\partial t}}=6x^2+4y+2xs^2+2ys^2$.
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{\bf Note:} It is acceptable to have $x,y$ featuring in the
answers as well as $t,s$. If you are asked to express
your answers {\bf only} in terms of the ``basic'' variables
$t,s$, then you would have to do an extra step: substituting 
everywhere you see $x$ or $y$ their expressions in terms of
$t$ and $s$. So in that case, the answers would be

$$
{{\partial z} \over {\partial s}}=3(s+2t)^2+2s^2t+4(s+2t)st+4(s^2t)st 
\quad ,
$$
$$
{{\partial z} \over {\partial t}}=6(s+2t)^2+4s^2t+2(s+2t)s^2+2(s^2t)s^2
\quad .
$$


{\bf  Do not bother to expand} (unless specifically asked to).


{\bf A Problem from a previous Final}

Find ${{\partial f} \over {\partial r}}$
and ${{\partial f} \over {\partial s}}$ as 
{\bf functions of $r$ and $s$ }, if
$$
f(x,y)= x^3 +2xy+ y^3 \quad,
$$
and the variables are related by $x=r-s$ and $y=r+s$.
You do not need to simplify!


{\bf Ans.} 
${{\partial f} \over {\partial r}}=3(r-s)^2+2(r+s)+3(r+s)^2+2(r-s)$, 
${{\partial f} \over {\partial s}}=-3(r-s)^2-2(r+s)+3(r+s)^2+2(r-s)$.

{\bf Another Problem from a Previous Final}

Find ${{\partial z} \over {\partial x}}$
and ${{\partial z} \over {\partial y}}$ if 
$$
\sin(x+2y+3z)=5xyz+1 \quad .
$$

{\bf Ans.} $(5yz-\cos(x+2y+3z))/(3\cos(x+2y+3z)-5xy)$;
$(5xz-2\cos(x+2y+3z))/(3\cos(x+2y+3z)-5xy)$ .

\end