<p>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:</p>
<p><span class="math-container">$$ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$</span></p>
<p>In this case:</p>
<p><span class="math-container">$$\dot{k} = \frac{d}{dt} \left(\frac{K(t)}{A(t)L(t)} \right)\\ = \frac{\dot{K}(t)A(t)L(t)-K(t)(\dot{A}L(t)+A(t)\dot{L}(t))}{(A(t)L(t))^2} \\= \frac{\dot{K}(t)A(t)L(t)}{(A(t)L(t))^2} - \frac{K(t)(\dot{A}(t)L(t)+A(t)\dot{L}(t))}{(A(t)L(t))^2}\\=  \frac{\dot{K}(t)}{(A(t)L(t))} - \frac{K(t)(\dot{A}(t)L(t)+A(t)\dot{L}(t))}{(A(t)L(t))^2}$$</span></p>
<p>So you just need to remember that <span class="math-container">$A$</span> and <span class="math-container">$L$</span> are both functions of <span class="math-container">$t$</span> 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)</p>

CC BY-SA 4.0
2020-12-06T18:47:03.490Z
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
Mr. T
CC BY-SA 4.0
2020-12-06T18:54:48.053Z
@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)
1muflon1
CC BY-SA 4.0
2020-12-06T19:16:45.400Z
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?
Mr. T
CC BY-SA 4.0
2020-12-06T19:40:04.597Z
@Mr.T you are not taking derivative with respect to K you are taking derivative with respect to t.
1muflon1
CC BY-SA 4.0
2020-12-06T19:48:53.290Z
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.
Mr. T
CC BY-SA 4.0
2020-12-06T19:49:33.973Z
@Mr.T sorry I meant to say denominator my bad. In denominator you have $A(t)$ and $L(t)$ so in the second term you need to apply product rule for those two
1muflon1
CC BY-SA 4.0
2020-12-06T17:02:51.197Z
2020-12-06T17:26:08.080Z
2020-12-06T17:26:08.080Z
1muflon1
<p>This is the beginning of the derivation of equation of motion from the Solow model (Romer, 2019):
<a href="https://i.sstatic.net/CGYq6.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CGYq6.jpg" alt="enter image description here" /></a></p>
<p>My question is: is this the total derivative (w.r.t. ultimate source of change &quot;t&quot;)? 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 &quot;+&quot; symbols. Thanks.</p>
<p>This is how I did it. Is it wrong? I did total derivative and got the same result as in the textbook:
<a href="https://i.sstatic.net/Vw1FL.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Vw1FL.jpg" alt="enter image description here" /></a></p>

<economic-growth><solow>
Is this a total derivative - if yes, why minus
Mr. T
# Is this a total derivative - if yes, why minus

(Asked by: username_0 on 2020-12-06)

This is the beginning of the derivation of equation of motion from the Solow model (Romer, 2019):
[![enter image description here](https://i.sstatic.net/CGYq6.jpg)](https://i.sstatic.net/CGYq6.jpg)

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](https://i.sstatic.net/Vw1FL.jpg)](https://i.sstatic.net/Vw1FL.jpg)

**Tags:** economic-growth, solow

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## 1 answers

### Answer by username_1 on 2020-12-06. Score: 2

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)

> **Comments:**
> **username_0**: 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  
> **username_1**: @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)  
> **username_0**: 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?  
> **username_1**: @Mr.T you are not taking derivative with respect to K you are taking derivative with respect to t.  
> **username_0**: 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.  
> **username_1**: @Mr.T sorry I meant to say denominator my bad. In denominator you have $A(t)$ and $L(t)$ so in the second term you need to apply product rule for those two  

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