# Thread: Differentiate and Solve for variable

1. ## Differentiate and Solve for variable

Question:

The function $Y=K^\frac{1}{3}L^\frac{2}{3}$ is Given.

a) Find $\frac{dY}{dL}$
b) Solve for "L" in the equation: $\frac{2}{3}K^\frac{1}{3}L^\frac{-1}{3}=\frac{W}{P}$

What I have so far:

a) $K^\frac{1}{3}\frac{2}{3}L^\frac{-1}{3}$
b) I am unsure, but would I would divide $\frac{W}{P}$ by K. Only thing is, I am sure you can't just ignore that exponent can you?

Help! I need a refresher. That much is obvious.

2. Originally Posted by mvho
Question:

The function $Y=K^\frac{1}{3}L^\frac{2}{3}$ is Given.

a) Find $\frac{dY}{dL}$
b) Solve for "L" in the equation: $\frac{2}{3}K^\frac{1}{3}L^\frac{-1}{3}=\frac{W}{P}$

What I have so far:

a) $K^\frac{1}{3}\frac{2}{3}L^\frac{-1}{3}$
b) I am unsure, but would I would divide $\frac{W}{P}$ by K. Only thing is, I am sure you can't just ignore that exponent can you?

Help! I need a refresher. That much is obvious.
HI

$\frac{2k^{-\frac{1}{3}}}{3L^{-\frac{1}{3}}}=\frac{W}{P}$

$3WL^{\frac{1}{3}}=2Pk^{\frac{1}{3}}$

$L^{\frac{1}{3}}=\frac{2Pk^{\frac{1}{3}}}{3W}$

Then cube both sides ..