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.


1 Answer 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, 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).

  • $\begingroup$ Thanks for answering this question, I will clarify the differention part in mathematics.SE then $\endgroup$
    – user30845
    Commented Oct 30, 2020 at 10:04
  • $\begingroup$ @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 $\endgroup$
    – 1muflon1
    Commented Oct 30, 2020 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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