# Thread: Trouble with exponents and differentiation

1. ## Trouble with exponents and differentiation

I think I have an error in my calculation, because I should get $s= \alpha$ at the end but I do not. Can someone please check to see where I've gone wrong?

Suppose you have: $c=(1-s) \left[ \frac{s}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$

Rewrite as

$
c=(1-s)s^\frac{\alpha}{1-\alpha} \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}
$

Expand the brackets:

$
c=(s^\frac{\alpha}{1-\alpha}-s^\frac{1}{1-\alpha}) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}
$

To find golden rule consumption, take the partial:

$
\frac{\partial c}{\partial s}=\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}
$

Set that beastie equal to zero so we can find the local max:

$
\frac{\partial c}{\partial s}=\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha} = 0
$

Divide through by $\left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$ to get

$
\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right)= 0
$

$
\frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha} = \frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha}
$

Multiply both sides by $1-\alpha$

$
\alpha s^\frac{2 \alpha -1}{1-\alpha} = s^\frac{\alpha}{1-\alpha}
$

Raise both sides to the power of $1-\alpha$

$
\alpha s^{2 \alpha -1} = s^{\alpha}
$

Multiply both sides by $s^{-\alpha}$

$
\alpha s^{\alpha -1} = 1
$

$
s^{\alpha -1} = \frac{1}{\alpha }
$

And finish by raising both sides by $\frac{1}{\alpha -1}$

$
s = \left( \frac{1}{\alpha } \right)^\frac{1}{\alpha -1}
$

We can rewrite as

$
s = \alpha^\frac{1}{1- \alpha}
$

But that hardly makes it look closer to $s=\alpha$, which I think is the right answer. What have I done wrong?

Originally Posted by garymarkhov
Raise both sides to the power of $1-\alpha$

$
\alpha s^{2 \alpha -1} = s^{\alpha}
$

When you raise a function ab^c to the power of d, you should get:
$
(\alpha b^c)^d = \alpha ^d*b^{cd}
$

but you have:
$
(\alpha b^c)^d = \alpha *b^{cd}
$

Note that the $\alpha$ term is missing an exponent.

3. Dear Lord,

Thank you for Mr. ebaines. Amen.

Love,

Gary