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\centerline
{
\bf Dr. Z's Math251 Handout \#14.3 [Partial Derivatives]
}

By Doron Zeilberger

{\bf Problem Type 14.3a}:  Find the first partial derivatives
of the function $f(x,y)$.

{\bf Example Problem 14.3a}: Find the first partial derivatives
of the function 
$$
f(x,y)=x^3y+ \ln(x+y) \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} When you take the partial derivative
with respect to $x$, then $x$ is the boss!,
and $y$ is to be treated just like a constant (number)
(it may even help, in your mind to replace $y$ by $y_0$
to bring home the fact that it is to be treated as a mere
constant, then at the end replace $y_0$ by $y$).

&
{\bf 1.} 
$$
{{\partial f} \over {\partial x}}=
(x^3y+ \ln(x+y))' \quad ,
$$
where at the {\it present context} the prime $'$ means
differentiation with respect to $x$.
So this equals
$$
3x^2 y+ {{1} \over {x+y}} \quad .
$$

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{\bf 2.} When you take the partial derivative w.r.t. to
$y$ it is the opposite: $y$ is the boss and $x$ is
treated as a constant.
&
{\bf 2.} 
$$
{{\partial f} \over {\partial y}}=
(x^3y+ \ln(x+y))' \quad ,
$$
where at the {\it present context} the prime $'$ means
differentiation with respect to $y$.
So this equals
$$
x^3 + {{1} \over {x+y}} \quad .
$$
{\bf Ans.}: ${{\partial f} \over {\partial x}}=3x^2 y+ {{1} \over {x+y}}$,
${{\partial f} \over {\partial y}}=x^3+ {{1} \over {x+y}} \,\,$ .
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{\bf Problem Type 14.3b}:  Use implicit differentiation
to find ${{\partial z} \over {\partial x}}$ and
${{\partial z} \over {\partial y}}$, given the
relationship 
$$
LeftSide(x,y,z)\, = \, RightSide(x,y,z) \quad .
$$
{\bf Example Problem 14.3b}: Use implicit differentiation
to find ${\partial z} \over {\partial x}$ and
${{\partial z} \over {\partial y}}$, given the
relationship 
$$
x^2+y^3+z^2=4xy^2z \quad.
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} When you take the partial derivative
with respect to $x$, then $x$ is the boss!,
and $y$ is to be treated just like a constant (number).
Also right now, $z$ is to be treated
as a function of x!
&
{\bf 1.} Differentiate with respect to $x$ the
given relation:
$$
(x^2+y^3+z^2)'=(4xy^2z)' \quad,
$$
where at the {\it present context} the prime $'$ means
differentiation with respect to $x$.
So this becomes
$$
2x+0+2zz'=(4y^2)[xz]' \quad,
$$
where we have taken $4y^2$ out since, in the present
context,  it is a {\bf constant}. By the
product rule applied to $[xz]'$,
$$
2x+2zz'=(4y^2)[x'z+xz']=(4y^2)[z+xz'] \quad.
$$
So much for calculus. We now have to use {\bf algebra} to
solve for $z'$ (i.e. ${{\partial z} \over {\partial x}}$).
Opening up parentheses,
$$
2x+2zz'=4y^2z+ 4y^2xz'  \quad.
$$
Moving all the terms involving $z'$ to the left and all the
other terms to the right yields
$$
2zz'-4y^2xz'=4y^2z -2x \quad.
$$
Factoring $z'$ out,
$$
(2z-4y^2x)z'=4y^2z -2x \quad.
$$
and finally, dividing by the term in front of $z'$,
to get $z'$ to be on its own, we get:
$$
z'={{4y^2z -2x} \over {2z-4y^2x}}
={{2y^2z -x} \over {z-2y^2x}} \quad .
$$
This is ${{\partial z} \over {\partial x}}$.

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{\bf 2.} When you take the partial derivative
with respect to $y$, then $y$ is the boss!,
and $x$ is to be treated just like a constant (number).
{\it Now} $z$ is to be treated as a function of y!
&
{\bf 2.} Differentiate with respect to $y$  the
given relation gives:
$$
(x^2+y^3+z^2)'=(4xy^2z)'
$$
where at the {\it present context} the prime $'$ means
differentiation with respect to $y$.
So this becomes
$$
0+3y^2+2zz'=(4x)[y^2z]' \quad,
$$
where we have taken $4x$ out since, in the present
context, it is a {\bf constant}. By the
product rule applied to $[y^2z]'$,
$$
3y^2+2zz'=(4x)[(y^2)'z+y^2 z']
$$
So
$$
3y^2+2zz'=(4x)[2yz+y^2 z']=8xyz+4xy^2z' \quad .
$$
So much for calculus. We have now to use {\bf algebra} to
solve for $z'$ (i.e. ${{\partial z} \over {\partial y}}$).
Moving all the terms involving $z'$ to the left and all the
other terms to the right yields
$$
2zz'-4xy^2z'=8xyz-3y^2 \quad.
$$
Factoring $z'$ out,
$$
(2z-4xy^2)z'=8xyz-3y^2 \quad,
$$
and finally, dividing by the term in front of $z'$,
to get $z'$ to be on its own, we get:
$$
z'={{8xyz-3y^2} \over {2z-4xy^2}}
={{(8xz-3y)y} \over {2(z-2xy^2)}} \quad .
$$
This is ${{\partial z} \over {\partial y}}$.
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