Elasticity Cobb-Douglas production function I am given the production function
$y=x_1^\alpha x_2^{1-\alpha}$, where $0&lt; \alpha &lt;1$ I found the demand functions for minimum production cost to be
$ x_1^{*}(w_1,w_2,y)=\left ( \frac{w_2}{w_1}\frac{\alpha}{\beta} \right )^{\frac{\beta}{\alpha +\beta}}y^\frac{1}{{\alpha +\beta}} \; \wedge \; x_2^{*}(w_1,w_2,y)=\left ( \frac{w_1}{w_2}\frac{\beta}{\alpha} \right )^{\frac{\alpha}{\alpha +\beta}}y^\frac{1}{{\alpha +\beta}}$
Problem
Now, I have to find the elasticity of $(x_2^*/x_1^*)$ wrt. $w_2/w_1$. I found that $(x_2^*/x_1^*)=\frac{\beta w_1}{\alpha w_2}$ and thus
$\epsilon = \frac{\partial \frac{\beta w_1}{\alpha w_2}}{\partial (w_2/w_1)} \cdot \frac{w_2/w_1}{\frac{\beta w_1}{\alpha w_2}}$

Here I end up stuck and do not know how to reduce/evaluate this expression
What does the elasticity mean for the total production cost on input 1?
For elasticity calculation why don't you try this:
\begin{align}
\frac{x_2^*}{x_1^*}&amp;=\frac{\beta w_1}{\alpha w_2}  \\
\ln\bigg(\frac{x_2^*}{x_1^*}\bigg) &amp;= c - \ln\bigg(\frac{w_2}{w_1}\bigg) \tag{$c=\ln(\beta/\alpha)$}
\end{align}
It's easy to see from above that elasticity is $-1$
For second part, cost of production:
\begin{align}
C&amp;=w_1x_1^* + w_2x_2^* \\
&amp;=w_1x_1^*(1+\beta/\alpha) \\
&amp;=\frac{w_1x_1^*}{\alpha} \tag{$\beta=1-\alpha$}
\end{align}
So the production cost per unit of input 1 is just the function of $w_1$ (given $\alpha$).
EDIT: As requested for clarification in comments:
Note that elasticity of $y$ w.r.t $x$ is defined as:
\begin{align}
\epsilon &amp;= \frac{\partial y/y}{\partial x/x} \\
&amp;= \frac{\partial \ln y}{\partial \ln x}
\end{align}
So elasticity is simply the slope on the log-scale for $y$ and $x$.
Now substitute $y=x_2^*/x_1^*$ and $x=w_2/w_1$ in your case.