# Ex: Find the Partial Derivatives of the Cobb Douglas Production Function

We’re given a Cobb-Douglas Production function P of L comma K. We want to find in the
marginal productivity of labor and the marginal productivity
of capital functions which would be the partial
derivative with respect to L and the partial derivative
with respect to K. Before we do this though, let’s talk more about the Cobb-Douglas
Production function. In economics, the Cobb-Douglas
Production function is a form of the production
function often used to represent the relationship between physical capital and labor and the amount of output that can be produced by those inputs. So this is the form of the
Cobb-Douglas Production function where P is the total production, L is the labor input,
K is the capital input A is the total factor productivity. And notice alpha and beta, the exponents, are the output and elasticity of labor and the output and elasticity of capital. So going back to our example, we’ll now find the marginal
productivity of labor function by determining the partial
derivative with respect to L. So to do this, we’ll
differentiate with respect to L treating K as a constant. So we’d have 12 times, if we differentiate with respect to L, we
would multiply by 0.7, and then we’d have L raised
to the power of 0.7 minus one, that’s negative 0.3 and we’re
treating K as a constant. Now let’s go ahead and simplify this and write this only
using positive exponents. So 12 times 0.7 is 8.4. So we’d have 8.4. K has a positive exponent, so that would be in the numerator. So K to the 0.3. And because L has a negative exponent, we’ll move it down to the denominator which would change the
sign of the exponent. So we’d have L to the positive
0.3 in the denominator. So this is the marginal
productivity of labor function. And then we’ll find the partial
derivative with respect to K which would be the marginal productivity of capital function. So we differentiate with respect to K, now treating L as a constant. So we’d have 12 and we
multiply by the exponent on K which would be 0.3. L will stay this same, L to the 0.7. And then for the exponent
on K, we would subtract one, 0.3 minus one is negative 0.7. Let’s go ahead and simplify. 12 times 0.3 is 3.6. L has a positive exponent. The denominator would be K
raised to the power of 0.7. Now that we have our two functions, we’re asked to evaluate them when L equals 400 and K equals 1,000. After we do this, we’ll
explain what this means. So the partial with respect to L, notice how the coordinates are L comma K. So it’d be 400 comma 1,000. Again, where L equals
400 and K equals 1,000. So for the partial with respect to L we would have 8.4 times K or 1,000 raised to the power of
0.3 divided by L or 400 to the 0.3. We’ll evaluate this in just a moment. Let’s go ahead and set up
the partial with respect to K at the same point. So we’ll have 3.6 times L or 400 raised to the power of 0.7 divided by K or 1,000 raised to the power of 0.7. Now we’ll evaluate
these on the calculator. So first we have in the numerator 8.4 and then times 1,000. This is raised to the power of .3, right arrow, closed parenthesis,
that’s the numerator, divided by 400 raised to the power of .3, Enter. If we round to three decimal places, this would be approximately 11.058. This is the change in production with respect to labor at this point. And next we have, and now for the partial with respect to K, we
have in the numerator 3.6 times 400 raised to the power of .7, right arrow, closed parenthesis,
there’s the numerator, divided by 1,000 raised to the power of .7, Enter. To three decimal places we’d
have approximately, 1.896. Now let’s emphasize what
these two values represent. The value from the
partial with respect to L which we found here is additional output that results from employing
an additional unit of labor when L equals 400 and K equals 1,000. So we can say at this point
production is increasing at approximately 11 units per worker. And then for the partial
with respect to K. This represents the additional output resulting from the use of an
additional unit of capital, again when L equals
400 and K equals 1,000. So the units on this value would be units of production per let’s say dollar. I hope you found this helpful.

## 5 thoughts on “Ex: Find the Partial Derivatives of the Cobb Douglas Production Function”

• #### Juni Tandra says:

Why isn't product rule being used here? Don't we treat K^0,3 as a function itself?

amazing

• #### Wisani Kenny says:

It was very helpful indeed. Thanks a lot.

• #### Aaron Barragan says:

thanks, i used this for my final project, stay smart!