Question on answering a derivative problem

[UDT89]

ok, i got class at 820, and i gotta leave my house in 5 hours.

The problem is

calculate dY/dK, dY/dL, and dY/dA, when Y=A*K^0.5*L^0.5


I am fairly familiar with the laws of derivatives, but forgot how to compute it with respect to K, L, and A


Anyone?

Gotta love macroeconomics. They're really making these last 7 credits hurt

 

[Darien]

ex: f(x,y) = x^3 + y^3

Differentiate with respect to y...
f (sub y) (x,y)..
--> x^3 + 3y^2

(in other words, your other terms become constants. In this case, x is held constant)

So to start you off,

Y(A, K, L) = (A) * (K^.5) * (L^.5)
dY/dK = (A) * (.5K^-.5) * (L^.5) [Derivative of K^.5 is .5K^-.5, A and L are held constant]

I'm 99% sure this is what it's supposed to be. Granted I haven't completed multivariable calculus, I was introduced to partial derivatives, gradiants and all sorts of nasty (and not so nasty) things in physics...

Hopefully I remember right

 

[UDT89]

thats what i did the first time, so im sure you're right............it just feels right

 

[optoman]

Aren't these partial derivatives? If they are then just take the derivative with respect to k for dy/dk, l for dy/dl and such.

Edit: damn I'm a slow typer. Darien is right on.

 

[FSUpaintball]

E=MC2?

 

[Darien]

Originally posted by: FSUpaintball
E=MC2?

At rest.

E(sub 0) = m(sub 0)C^2

E^2 = (p^2)(c^2) + ((m(sub 0)^2)(c^2))^2; E = K + E (sub 0)



(It's 11:05PM...and while I should be reading chemistry, I'm posting on AT. Will take a nap and go to this subject later. I hate chemistry. Gimme the integral over balancing equations and finding out "stuff"!)