<p>They used a product rule, quotient rule and chain rule, I won't be discussing the rules themselves  as such topics belong to <a href="https://math.stackexchange.com/">Mathematics.SE</a>, but the reason why you might not realize this outright is the particular notation in a growth theory, which would be on topic here so I will focus on that.</p>
<p>In Solow growth model with labor augmenting technological change <span class="math-container">$f(k)$</span> is the production function expressed per 'effective worker' so:</p>
<p><span class="math-container">$$f(k) = F \left(\frac{K}{AL},1 \right)$$</span></p>
<p>so you might as well rewrite the problem as:</p>
<p><span class="math-container">$$d\left(AL F \left(\frac{K}{AL},1 \right) \right)/dL=A F \left(\frac{K}{AL},1 \right) +AL F '\left(\frac{K}{AL},1 \right) (−AK)/(LA)^2$$</span>
Now we can cancel some terms and use the fact that <span class="math-container">$k =K/AL$</span>:</p>
<p><span class="math-container">$$A F \left(\frac{K}{AL},1 \right) + F' \left(\frac{K}{AL},1 \right) (−K)/(AL)=A F \left(\frac{K}{AL},1 \right)  -A k F' \left(\frac{K}{AL},1 \right)  $$</span></p>
<p>Now finally go replace the <span class="math-container">$ F \left(\frac{K}{AL},1 \right)$</span> back with <span class="math-container">$f(k)$</span> so you get:</p>
<p><span class="math-container">$$A F \left(\frac{K}{AL},1 \right) - Ak F' \left(\frac{K}{AL},1 \right) =A f(k)  -Ak f'(k) = A(f(k)  -k f'(k)) $$</span></p>
<p>So it is all about paying attention to what different definition of the arguments used here are. If you can't follow the differentiation itself that is question for Mathematics.SE or just further self study (Essential Mathematics for Economic Analysis by Sydsaeter covers all you need for this sort of problem).</p>

CC BY-SA 4.0
2020-10-30T10:04:54.240Z
Thanks for answering this question, I will clarify the differention part in mathematics.SE then
user30845
CC BY-SA 4.0
2020-10-30T11:26:04.140Z
@Joker312 you are welcome also I made small mistake it was actually supposed to be $\frac{K}{L^2A}$ not $\frac{K}{(LA)^2}$ (actually the term would be $\frac{AK}{(LA)^2}$ but A's in numerator and denominator cancel). However, everything else stays the same, and I edited the answer to reflect that
1muflon1
CC BY-SA 4.0
2020-10-30T09:43:13.043Z
2020-10-30T11:24:37.560Z
2020-10-30T11:24:37.560Z
1muflon1
<p>I am taking macro course this Fall and my calculus is quite rusty. So in the lecture notes they derive the following:
<span class="math-container">$$MPL = dY/dL = d(ALf(k))/dL = Af(k) + ALf'(k)  (-K)/(L^2  A) = A(f(k) - kf'(k)) = w$$</span></p>
<p>Specifically, I don't quite understand how they got <span class="math-container">$Af(k) + ALf'(k) (-K)/(L^2  A)$</span> from <span class="math-container">$d(ALf(k))/dL?$</span></p>
<p>Honestly, I am stuck at this point and guess will be doing a lot of differentiation like this during the entire course.</p>

<macroeconomics><solow>
How to take the derivative and show that MPL in the Solow model is equal to real wage w
user30845
# How to take the derivative and show that MPL in the Solow model is equal to real wage w

(Asked by: username_0 on 2020-10-30)

I am taking macro course this Fall and my calculus is quite rusty. So in the lecture notes they derive the following:
$$MPL = dY/dL = d(ALf(k))/dL = Af(k) + ALf'(k) (-K)/(L^2 A) = A(f(k) - kf'(k)) = w$$

Specifically, I don't quite understand how they got $Af(k) + ALf'(k) (-K)/(L^2 A)$ from $d(ALf(k))/dL?$

Honestly, I am stuck at this point and guess will be doing a lot of differentiation like this during the entire course.

**Tags:** macroeconomics, solow


---

## 1 answers

### Answer by username_1 on 2020-10-30. Score: 1

They used a product rule, quotient rule and chain rule, I won't be discussing the rules themselves as such topics belong to [Mathematics.SE](https://math.stackexchange.com/), but the reason why you might not realize this outright is the particular notation in a growth theory, which would be on topic here so I will focus on that.

In Solow growth model with labor augmenting technological change $f(k)$ is the production function expressed per 'effective worker' so:

$$f(k) = F \left(\frac{K}{AL},1 \right)$$

so you might as well rewrite the problem as:

$$d\left(AL F \left(\frac{K}{AL},1 \right) \right)/dL=A F \left(\frac{K}{AL},1 \right) +AL F '\left(\frac{K}{AL},1 \right) (−AK)/(LA)^2$$
Now we can cancel some terms and use the fact that $k =K/AL$:

$$A F \left(\frac{K}{AL},1 \right) + F' \left(\frac{K}{AL},1 \right) (−K)/(AL)=A F \left(\frac{K}{AL},1 \right) -A k F' \left(\frac{K}{AL},1 \right) $$

Now finally go replace the $ F \left(\frac{K}{AL},1 \right)$ back with $f(k)$ so you get:

$$A F \left(\frac{K}{AL},1 \right) - Ak F' \left(\frac{K}{AL},1 \right) =A f(k) -Ak f'(k) = A(f(k) -k f'(k)) $$

So it is all about paying attention to what different definition of the arguments used here are. If you can't follow the differentiation itself that is question for Mathematics.SE or just further self study (Essential Mathematics for Economic Analysis by Sydsaeter covers all you need for this sort of problem).

> **Comments:**
> **username_0**: Thanks for answering this question, I will clarify the differention part in mathematics.SE then  
> **username_1**: @Joker312 you are welcome also I made small mistake it was actually supposed to be $\frac{K}{L^2A}$ not $\frac{K}{(LA)^2}$ (actually the term would be $\frac{AK}{(LA)^2}$ but A's in numerator and denominator cancel). However, everything else stays the same, and I edited the answer to reflect that  

---

