Spelling Out Kathy’s Unit 6 Tamar ZEILBERGER

The Utility is

U(x) = x^{α} (Z − x)^{1−α} .

We need to find the first-derivative if U(x), called U′(x). U(x) is a product of two simpler functions, x^{α} and (Z − x)^{1−α}, so we need the product rule from calc1. Recall that the product rule is:

(f(x)g(x))′ = f′(x)g(x) + f(x)g′(x) .

Taking f(x) = x^{α} and g(x) = (Z − x)^{1−α}, we have

U′(x) = (x^{α} (Z − x)^{1−α})′ = (x^{α})′ (Z − x)^{1−α} + x^{α} ((Z − x)^{1−α})′ .  (1)

Recall from calc1 that (x^{n})′ = nx^{n−1} for any n, so (x^{α})′ = αx^{α−1} .

Also recall from the chain rule

f(g(x))′ = f′(g(x)) g′(x)

So ((Z−x)^{1−α})′ = (1−α)(Z−x)^{1−α−1}·(Z−x)′ = (1−α)(Z−x)^{1−α−1}·(−1) = (1−α)(Z−x)^{−α}·(−1) = −(1−α)(Z−x)^{−α} =

Going back to (1) we have U′(x) = αx^{α−1} (Z − x)^{1−α} − (1 − α)x^{α} ((Z − x)^{−α} .

Note: Kathy had a typo she had ‘+’ rather than the correct ‘−’ in the middle. Factoring out x^{α−1}(Z − x)^{−α} we get

U′(x) = x^{α−1}(Z − x)^{−α}(α(Z − x) − (1 − α)x)) .

We have to solve U′(x) = 0. Dividing by the front x^{α−1}(Z − x)^{−α} we have

α(Z − x) − (1 − α)x = 0 .

Simplifying

αZ − αx − x + αx = 0 .

Simplifying more

αZ − x = 0 .

So x = αZ. This is our critical point The rest is a tedious U′′(x).