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%%a Plain TeX file by Shalosh B. Ekhad and Doron Zeilberger (x pages)

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\centerline
{\bf  Spelling Out Kathy's Unit 6 }
\bigskip
\centerline
{\it Tamar ZEILBERGER}


The {\bf Utility} is
$$
U(x)= x^\alpha \, (Z- x)^{1-\alpha} \quad.
$$

We need to find the {\bf first-derivative} if $U(x)$, called $U'(x)$.

$U(x)$ is a {\bf product} of two simpler functions, $x^{\alpha}$ and  $(Z- x)^{1-\alpha}$, so we need the
 {\bf product rule} from calc1.

Recall that the product rule is:
$$
(f(x) g(x) )' = f'(x) g(x)  \, + \, f(x) g'(x) \quad.
$$


Taking $f(x)=x^{\alpha}$ and  $g(x)=(Z- x)^{1-\alpha}$, we have
$$
U'(x)=( x^\alpha \, (Z- x)^{1-\alpha} )' =  (x^\alpha)' \, (Z- x)^{1-\alpha} +  x^\alpha \, ( (Z- x)^{1-\alpha})'  \quad.
\eqno(1)
$$


Recall from calc1 that $(x^n)'=n x^{n-1}$ for {\bf any} $n$, so
$$
 (x^\alpha)'= \alpha x^{\alpha -1} \quad .
$$
\bigskip
Also recall from the {\bf chain rule}
$$
f(g(x))' =f'(g(x)) \, g'(x)
$$

So
$$
 ( (Z- x)^{1-\alpha})'= (1-\alpha) (Z- x)^{1-\alpha-1} \cdot (Z-x)' =
(1-\alpha) (Z- x)^{1-\alpha-1} \cdot (-1) =
(1-\alpha) (Z- x)^{-\alpha} \cdot (-1) =
-(1-\alpha) (Z- x)^{-\alpha} =
$$

Going back to $(1)$ we have
$$
U'(x)=  \alpha x^{\alpha -1} \, (Z- x)^{1-\alpha} \, - \, (1- \alpha)  x^\alpha \, ( (Z- x)^{-\alpha}  \quad.
$$

{\bf Note:} Kathy had a typo she had `$+$' rather than the correct `$-$' in the middle.

Factoring out  $x^{\alpha -1} (Z-x)^{-\alpha}$ we get
$$
U'(x)=  x^{\alpha -1} (Z-x)^{-\alpha} (  \alpha (Z-x) \, - \, (1- \alpha)  x ) ) \quad.
$$
We have to solve $U'(x)=0$. Dividing by  the front $x^{\alpha -1} (Z-x)^{-\alpha}$ we have
$$
 \alpha (Z-x) \, - \, (1- \alpha)  x  \, = \, 0 \quad.
$$
Simplifying
$$
 \alpha Z- \alpha x -x + \alpha  x  \, = \, 0 \quad.
$$
Simplifying more
$$
 \alpha Z -  x   \, = \, 0 \quad.
$$
So $x=\alpha Z$. This is our {\bf critical point}

The rest is a tedious $U''(x)$.



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