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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

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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)

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  • $\begingroup$ 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

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