The problem asks for the derivative $\frac{dK}{dL}$ from the equation:

$1000 = 1 \cdot L^{0.5} \cdot K^{0.5}.$

This equation resembles a Cobb-Douglas production function, where $L$ and $K$ are inputs, raised to exponents of 0.5.

Let’s solve for $\frac{dK}{dL}$ step-by-step.

Step 1: Express $K$ in terms of $L$

We start by isolating $K$ in the equation:

$L^{0.5} \cdot K^{0.5} = 1000.$

Divide both sides by $L^{0.5}$:

$K^{0.5} = \frac{1000}{L^{0.5}}.$

Square both sides to remove the exponent on $K$:

$K = \left( \frac{1000}{L^{0.5}} \right)^2.$

Simplify the right-hand side:

$K = \frac{1000^2}{L} = \frac{1,000,000}{L}.$

Step 2: Differentiate $K$ with respect to $L$

Now we apply the power rule for differentiation. The equation we have is:

$K = 1,000,000 \cdot L^{-1}.$

The derivative of $K$ with respect to $L$ is:

$\frac{dK}{dL} = 1,000,000 \cdot (-1) \cdot L^{-2}.$

Simplify:

$\frac{dK}{dL} = -\frac{1,000,000}{L^2}.$

Final Answer

The derivative $\frac{dK}{dL}$ is:

$\boxed{-\frac{1,000,000}{L^2}}.$

Would you like further details on any step?

Related Questions:

1. What is the Cobb-Douglas production function and how is it used in economics?
2. How do we perform implicit differentiation with multiple variables?
3. What are the rules for differentiating functions with negative exponents?
4. How do elasticity concepts relate to the Cobb-Douglas function?
5. How does changing exponents affect the shape of production functions?

Tip: When differentiating a product or quotient involving exponents, rewrite the expression in a simpler form first. It makes differentiation much easier!