# derivatives with a twist

• Jan 21st 2009, 06:15 PM
foxracing
derivatives with a twist
Hey, just found this forum and wanted to give a quick thanks beforehand.

I took calc a year ago and did ok, but now i need to know some calc for an econ class im in.

Here's a sample question...
Q=10L^.75 K^.25
In my calc class we just had to do the math, and i would multiply the exponent to the coefficient and minus the exponent by 1. I'm 99.9% this is correct.

But my econ teacher wants us to caclucate to sets:
a. calculate derivative Q/derivative L
b. calculate derivative Q/derivative K

I'm lost on that part. Thanks
• Jan 21st 2009, 06:45 PM
mollymcf2009
Is the sample equation you posted the equation you are supposed to use to determine the two sets for a value? You will need to use the product rule to get the derivative for the sample problem because you are multiplying two variables and the quotient rule for the Q/K problems. Can you be more specific about what you are solving for?
• Jan 21st 2009, 06:55 PM
foxracing
Sorry if i wasnt too clear.
Heres the exact problem:

Consider the following function Q= 10L^.75 K^.25
a. Calculate derivative Q/derivative L
b. Calculate derivative Q/derivative k

I figure if i can figure out a or b, i can do the other one.
• Jan 21st 2009, 07:22 PM
mollymcf2009
Product rule:
Q = lk
Q' = l'k +lk'

So,

Q = $10l^{.75}k^{.25}$

Q' = $7.5l^{-.25}k^{.25} + 10l^{.75}.25k^{-.75}$

I laid out the product rule for you. See if you can finish it and then go from there. Come back if you get stuck!!
• Jan 21st 2009, 09:27 PM
mr fantastic
Quote:

Originally Posted by foxracing
Sorry if i wasnt too clear.
Heres the exact problem:

Consider the following function Q= 10L^.75 K^.25
a. Calculate derivative Q/derivative L
b. Calculate derivative Q/derivative k

I figure if i can figure out a or b, i can do the other one.

The question as you've posted it requires the idea of a partial derivative. have you learned about this?

To get $\frac{\partial Q}{\partial L}$, you treat K as if it were a constant. Then:

$\frac{\partial Q}{\partial L} = 0.75 L^{-0.25} K^{0.25}$.

The other is done in a similar way and is left for you to attempt.
• Jan 21st 2009, 09:29 PM
mr fantastic
Quote:

Originally Posted by mollymcf2009
Product rule:
Q = lk
Q' = l'k +lk'

So,

Q = $10l^{.75}k^{.25}$

Q' = $7.5l^{-.25}k^{.25} + 10l^{.75}.25k^{-.75}$

I laid out the product rule for you. See if you can finish it and then go from there. Come back if you get stuck!!

The problem as posted requires the idea of a partial derivative. See my post above. So gettting a total derivative (with respect to x ....?) is not correct.
• Jan 21st 2009, 09:34 PM
Mush
Quote:

Originally Posted by foxracing
Hey, just found this forum and wanted to give a quick thanks beforehand.

I took calc a year ago and did ok, but now i need to know some calc for an econ class im in.

Here's a sample question...
Q=10L^.75 K^.25
In my calc class we just had to do the math, and i would multiply the exponent to the coefficient and minus the exponent by 1. I'm 99.9% this is correct.

But my econ teacher wants us to caclucate to sets:
a. calculate derivative Q/derivative L
b. calculate derivative Q/derivative K

I'm lost on that part. Thanks

$\frac{dQ}{dK}$ means you differentiate the equation with respect to variable K, and present that all other variables are constants! $\frac{dQ}{dL}$ means the same thing, except you differentiate with respect to L, and present that K is a constant!

SO!

$\frac{dQ}{dK} = \frac{d}{dQ} 10L^{\frac{3}{4}} K^{\frac{1}{4}}$

$= 10L^{\frac{3}{4}}\frac{d}{dQ} (K^{\frac{1}{4}})$

$= 10L^{\frac{3}{4}} ( \frac{1}{4}K^{\frac{-3}{4}})$

$= \frac{10}{4}L^{\frac{3}{4}} K^{\frac{-3}{4}}$

$= \frac{5}{2}L^{\frac{3}{4}} K^{\frac{-3}{4}}$

$\frac{dQ}{dL} = \frac{d}{dL} 10L^{\frac{3}{4}} K^{\frac{1}{4}}$

$= K^{\frac{1}{4}} \frac{d}{dQ}(10L^{\frac{3}{4}})$

$= K^{\frac{1}{4}} (\frac{3}{4}10L^{\frac{-1}{4}})$

$= K^{\frac{1}{4}} (\frac{30}{4}L^{\frac{-1}{4}})$

$= K^{\frac{1}{4}} (\frac{15}{2}L^{\frac{-1}{4}})$

$= \frac{15}{2}K^{\frac{1}{4}} L^{\frac{-1}{4}}$