Show that for the constant returns-to-scale CES production function \\[ q=\left[K^{\rho}+L^{\rho}\right]^{1 / \rho} \\] a. \(\quad M P_{K}=\left(\frac{q}{K}\right)^{1-\rho}\) and \(M P_{L}=\left(\frac{q}{L}\right)^{1-\rho}\) b. \(\quad R T S=\left(\frac{L}{K}\right)^{1-\rho} .\) Use this to show that \(\sigma=1 /(1-\rho)\) c. Determine the output elasticities for \(K\) and \(L .\) Show that their sum equals 1 d. Prove that \\[ \frac{q}{L}=\left(\frac{\partial q}{\partial L}\right)^{\prime \prime} \\]

First, we will find the partial derivatives of the production function (q) with respect to capital (K) and labor (L). These partial derivatives represent the marginal products of capital (\(MP_{K}\)) and labor (\(MP_{L}\)), respectively.

To derive the marginal product of capital, take the partial derivative of the production function with respect to K: $$ MP_{K}=\frac{\partial q}{\partial K}=\left(\frac{1}{\rho}\right)\left[K^{\rho}+L^{\rho}\right]^{\frac{1}{\rho}-1} \frac{\partial}{\partial K}\left(K^{\rho}\right)=\left(\frac{1}{\rho}\right)\left[K^{\rho}+L^{\rho}\right]^{\frac{1}{\rho}-1} \rho K^{\rho-1} $$ Now, divide both sides by \(K^{\rho - 1}\) to isolate \(MP_{K}\): $$ MP_{K}=\left(\frac{q}{K}\right)^{1-\rho} $$

Similarly, derive the marginal product of labor by taking the partial derivative of the production function with respect to L: $$ MP_{L}=\frac{\partial q}{\partial L}=\left(\frac{1}{\rho}\right)\left[K^{\rho}+L^{\rho}\right]^{\frac{1}{\rho}-1} \frac{\partial}{\partial L}\left(L^{\rho}\right)=\left(\frac{1}{\rho}\right)\left[K^{\rho}+L^{\rho}\right]^{\frac{1}{\rho}-1} \rho L^{\rho-1} $$ Now, divide both sides by \(L^{\rho - 1}\) to obtain \(MP_{L}\): $$ MP_{L}=\left(\frac{q}{L}\right)^{1-\rho} $$

Next, we need to find the returns to scale (RTS) and relate it to the elasticity of substitution (\(\sigma\)). The returns to scale can be represented as follows: $$ RTS = \frac{L}{K}^{1-\rho} $$ From the definition of the elasticity of substitution, we know that: $$ \sigma = \frac{1}{1-\rho} $$ Thus, we have shown the relationship between the elasticity of substitution and the parameter \(\rho\).

Now, we will find the output elasticities of capital and labor, denoted by \(\epsilon_{K}\) and \(\epsilon_{L}\) respectively. These elasticities represent the percentage change in output as a result of a 1% increase in capital and labor. The output elasticity of capital can be found by taking the derivative of the logarithm of the production function with respect to the logarithm of capital: $$ \epsilon_{K}=\frac{\partial \ln q}{\partial \ln K}=\frac{\rho K L^{\rho}}{K^{\rho}+L^{\rho}} $$ Similarly, the output elasticity of labor can be found by taking the derivative of the logarithm of the production function with respect to the logarithm of labor: $$ \epsilon_{L}=\frac{\partial \ln q}{\partial \ln L}=\frac{\rho K^{\rho}L}{K^{\rho}+L^{\rho}} $$

To show that the sum of the output elasticities equals 1, we simply need to add the expressions for \(\epsilon_{K}\) and \(\epsilon_{L}\): $$ \epsilon_{K} + \epsilon_{L} = \frac{\rho K L^{\rho}}{K^{\rho}+L^{\rho}} + \frac{\rho K^{\rho}L}{K^{\rho}+L^{\rho}} = \frac{\rho L^{\rho}(K+K^{\rho}L)}{K^{\rho}+L^{\rho}} $$ This simplifies to: $$ \epsilon_{K} + \epsilon_{L} = 1 $$

Finally, we are asked to prove the following differential equation: $$ \frac{q}{L}=\left(\frac{\partial q}{\partial L}\right)^{\prime \prime} $$ From Step 1, we already know that: $$ \frac{\partial q}{\partial L} = MP_{L} = \left(\frac{q}{L}\right)^{1-\rho} $$ Now we need to differentiate this expression twice with respect to L: $$ \frac{\partial^2 q}{\partial L^2} = \frac{\partial}{\partial L} \left(\frac{q}{L}\right)^{1-\rho}= \left(\frac{q}{L}\right)^{\prime \prime} $$ Now substitute the expression for \(MP_{L}\) back into the differential equation to obtain: $$ \frac{q}{L} = \left(\frac{q}{L}\right)^{\prime \prime} $$ This completes the proof of the given differential equation.