0
$\begingroup$

This is the beginning of the derivation of equation of motion from the Solow model (Romer, 2019): enter image description here

My question is: is this the total derivative (w.r.t. ultimate source of change "t")? If yes, why there is a minus sign in front of the 2nd and 3rd term? I assume because they are in the denominator. But I was not able find any rules for this on net. Everywhere are only "+" symbols. Thanks.

This is how I did it. Is it wrong? I did total derivative and got the same result as in the textbook: enter image description here

Improve this question
$\endgroup$
1
$\begingroup$

This is just standard derivative. The minus sign there is because you are taking derivative of a quotient and the quotient rule for derivatives is:

$$ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$

In this case:

$$\dot{k} = \frac{d}{dt} \left(\frac{K(t)}{A(t)L(t)} \right)\\ = \frac{\dot{K}(t)A(t)L(t)-K(t)(\dot{A}L(t)+A(t)\dot{L}(t))}{(A(t)L(t))^2} \\= \frac{\dot{K}(t)A(t)L(t)}{(A(t)L(t))^2} - \frac{K(t)(\dot{A}(t)L(t)+A(t)\dot{L}(t))}{(A(t)L(t))^2}\\= \frac{\dot{K}(t)}{(A(t)L(t))} - \frac{K(t)(\dot{A}(t)L(t)+A(t)\dot{L}(t))}{(A(t)L(t))^2}$$

So you just need to remember that $A$ and $L$ are both functions of $t$ so you need to take derivative of whole qutient and quotient rule has minus there (if you want explanation for why quotient rule includes minus the right place is mathematics.se)

Improve this answer
$\endgroup$
6
  • $\begingroup$ This is how I did it. Is it wrong? I did total derivative and got the same result as in the textbook - I was not able to add an image here, so I am sticking it to my original post $\endgroup$ – Mr. T Dec 6 '20 at 18:47
  • $\begingroup$ @Mr.T there is a mistake because for the numerator you also need to apply the product rule since $d/dx[f(x)g(x) ] = f'(x)g(x) + f(x)g'(x)$, there might be also some other mistakes but I cant read the handwriting please post it using mathjax (the same way as I posted equations in my answer) $\endgroup$ – 1muflon1♦ Dec 6 '20 at 18:54
  • $\begingroup$ I will try to upload it using mathjax, but I need to get used to it, sorry for handwriting. Why do I need to apply a product rule to the numerator, as there is only a single K? $\endgroup$ – Mr. T Dec 6 '20 at 19:16
  • $\begingroup$ @Mr.T you are not taking derivative with respect to K you are taking derivative with respect to t. $\endgroup$ – 1muflon1♦ Dec 6 '20 at 19:40
  • $\begingroup$ Yes, but there is no multiplication. In the numerator is K(t), K as a function of time. It is not multiplication, it is a composite function. Hence chain rule should be used, I dont see why the product rule should be used. $\endgroup$ – Mr. T Dec 6 '20 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.