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Once i saw a TA doing this, but it seems a little bit weird for me

$$ \frac{dF(KL)}{d(KL)}=\frac{L*dF(K)}{L*d(K)}=\frac{dF(K)}{d(K)} $$

Is it possible to asume this for any homogeneous function of degree 1? Or is this due to the chain rule?

Thanks in advance )

mvw
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NoseNada
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1 Answers1

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A little weird indeed ,You know what kind of notation it is ;BAD notation ,probably a short hand for researchers in a hurry . Lets see F is a function ,lets put y =F(x) (y is a function of X) . Let x =KL L a fixed nonzero constant throughout (x is a function of K ) . Let K be an unspecified real number ( the letter"K" is called of course a variable ).Composing we get y= F(LK) so y is also a function of K (it's name is NOT F) .

The chain rule says dy/dK =dy/dx * dx/dK =dy/dx*L We can write this loosely as

(1/L) d F(LK)/dK = d F(LK)/d(LK) .Now is f is homogeneous then F(LK) =LF(K) and since d LF(K)/dK= LdF(K)/dK the L and 1/L cancel giving dF(K)/dK=dF(LK)/(LK) . So yes you get the right answer for homogeneous functions .Guess the notation can be useful even if bad .Regards ,Stuart M.N.

user439545
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  • I think you should use Latex in your answer to make it readable. – Jannik Pitt May 08 '17 at 09:10
  • What is Latex ? – user439545 May 08 '17 at 22:43
  • That was quite useful, I'm going to rewrite it just for the sake of it: $\frac{dF(KL)}{d(K)}=\frac{dF(KL)}{d(KL)}\frac{dKL}{dK}=\frac{dF(KL)}{d(KL)}L \leftrightarrow \frac{dF(KL)}{d(K)} \frac{1}{L}=\frac{dF(KL)}{d(KL)}$ Since $y$ is an homogeneous function of degree "1": $F(KL)=LF(K)$. Thus: $\frac{dF(KL)}{d(KL)}=\frac{dF(K)}{d(K)}$ Easy peasy lemon squeezy. – NoseNada May 09 '17 at 15:03