# Thread:

1. ## partial derivative without knowing actual function?

My textbook, Advanced Macroeconomics (Romer), introduces a function $Y=F(K,AL)$, which can be written $\frac{Y}{AL}=F(\frac{K}{AL},1)$ and is often shortened to $y=f(k)$. It then suggests that $\frac{\partial F(K,AL)}{\partial K} = ALf'(\frac{K}{AL})(\frac{1}{AL})$.

But how can they take the partial with respect to K if they don't specify the actual function? They say what the function depends on, but not what the actual function is. What is happening here?

2. Looks like they used the Chain Rule:

$\frac{\partial}{\partial K}\left(AL\cdot F\left(\frac{K}{AL},1\right)\right)=AL\frac{\parti al F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{\partial}{ \partial K}\left(\frac{K}{AL}\right)=AL\frac{\partial F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{1}{AL}.$

3. Originally Posted by Scott H
Looks like they used the Chain Rule:

$\frac{\partial}{\partial K}\left(AL\cdot F\left(\frac{K}{AL},1\right)\right)=AL\frac{\parti al F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{\partial}{ \partial K}\left(\frac{K}{AL}\right)=AL\frac{\partial F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{1}{AL}.$
That looks about right. A little strange to find the partial without knowing what the actual function is, but I guess it would work out if we plugged in the actual function?

4. Correct, as the Chain Rule applies to every function.

5. Originally Posted by Scott H
Correct, as the Chain Rule applies to every function.
Hmm, so say you discovered that the function is $Y=F(K,AL)=K^\alpha(AL)^{1-\alpha}$. Can we figure out what the partial with respect to K is from $f'(\frac{K}{AL})$?

6. Yes.

Differentiating the function normally, we obtain

$\frac{\partial}{\partial K}(K^{\alpha}(AL)^{1-\alpha})=\alpha K^{\alpha-1}(AL)^{1-\alpha}.$

Using the formula derived from the Chain Rule, we obtain

$\begin{aligned}<br /> \frac{\partial}{\partial K}\left(AL\cdot F\left(\frac{K}{AL},1\right)\right)&=AL\frac{\part ial F}{\partial K}\left(\frac{K}{AL},1\right)\cdot\frac{1}{AL}\\<br /> &=AL\alpha\left(\frac{K}{AL}\right)^{\alpha-1}(1)^{1-\alpha}\cdot\frac{1}{AL}\\<br /> &=\frac{\alpha K^{\alpha-1}}{(AL)^{\alpha-1}}\\<br /> &=\alpha K^{\alpha-1}(AL)^{1-\alpha}.<br /> \end{aligned}<br />$

7. That's great, thanks for your help!
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