Question

A partial derivative question implementing production function model.

Answer 1

I've tried everything but cant figure out how my professor arrived at that answer.

Answer 2

$$
Y_K={\partial\over \partial K}(A K^a+B L^a) = a A K^{a-1}\\
Y_L={\partial\over \partial L}(A K^a+B L^a) = a B L^{a-1}\\
aY=K\times a A K^{a-1}+L\times a B L^{a-1}=KY_K+LY_L
$$
This might help - http://www.wolframalpha.com/input/?i=d%2Fdx+%28A*x%5Ea%2BB*L%5Ea%29

Answer 3

I got the Yk YL or the first two lines, but hte third line could you explain why you used aY?

Answer 4

Let me add a few other steps
$$
Y_K={\partial\over \partial K}(A K^a+B L^a) = a A K^{a-1}\\
Y_L={\partial\over \partial L}(A K^a+B L^a) = a B L^{a-1}\\
KY_K+LY_L=K\times a A K^{a-1}+L\times a B L^{a}=aAK^{a}+aBL^a\\
=a(AK^{a}+BL^a)=aY
$$

Answer 5

I got the Yk YL or the first two lines, but hte third line could you explain why you used aY?

Answer 6

We are given
$$
Y=AK^a+BL^a
$$
Everything else follows from this

Answer 7

a(AKa+BLa)=aY

Could you explain this step?

Answer 8

Do see how
$$
aAK^{a}+aBL^a
=a(AK^{a}+BL^a)
$$
?

Answer 9

I get a(AKa+BLa) why do you put = aY

Answer 10

Because
$$
Y=AK^{a}+BL^a\\
\text{so}\\
a(AK^{a}+BL^a)=aY
$$

Answer 11

Ohhh wow I completely missed that thanks so much for explaining!

Answer 12

Im gonna solve it again just for practice haha :)

Answer 13

No problem :)

You're welcome!

Answer 14

That's actually a VERY good idea!