partial derivative without knowing actual function?

**garymarkhov** · October 16th 2009, 02:50 PM

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?

**Scott H** · October 16th 2009, 03:09 PM

Looks like they used the Chain Rule:

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

**garymarkhov** · October 16th 2009, 03:32 PM

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?

**Scott H** · October 16th 2009, 03:44 PM

Correct, as the Chain Rule applies to every function.

**garymarkhov** · October 16th 2009, 04:00 PM

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 $f'(\frac{K}{AL})$ ?

**Scott H** · October 17th 2009, 08:41 AM

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} \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}\\ &=AL\alpha\left(\frac{K}{AL}\right)^{\alpha-1}(1)^{1-\alpha}\cdot\frac{1}{AL}\\ &=\frac{\alpha K^{\alpha-1}}{(AL)^{\alpha-1}}\\ &=\alpha K^{\alpha-1}(AL)^{1-\alpha}. \end{aligned} $

**garymarkhov** · October 23rd 2009, 03:07 AM

That's great, thanks for your help!

Thread: partial derivative without knowing actual function?

LinkBack

Thread Tools

Display

partial derivative without knowing actual function?

Similar Math Help Forum Discussions

Need help with a partial derivative of a multivariate function

Finding the equation of a derivative line wihtout knowing coordinates

Partial derivative of a function with exp in it

Partial derivative of a function

Second partial derivative of this function at (0,0)

Search Tags