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I am trying to solve all steps in a economics paper , but after spending two days with the same differentiation Im losing faith. Can someone out there help me? The problem:

Differentiate: $$ \begin{align} &R_{pg}(w) = \left(\frac{r + \mu + \delta_g}{r + \mu + \delta_p}\right)w \\ &+\left( \left(\frac{r + \mu + \delta_g}{r + \mu + \delta_p}\right)\lambda_{pp} - \lambda_{gp}\right) \int_{w}^{\infty} W'_{p}(x)(1 - F_{p}(x)) \, dx \\ &+\left( \left(\frac{r + \mu + \delta_g}{r + \mu + \delta_p}\right)\lambda_{pg} - \lambda_{gg}\right) \int_{R_{pg}(w)}^{\infty} W'_{g}(x)(1 - F_{g}(x)) \, dx \end{align} $$

additional info: $$ W'_{p}(x) = \left[r + \mu + \delta_{p} + \lambda_{pp} (1 - F_{p}(w)) + \lambda_{pg} (1 - F_{g}(R_{pg}(w)))\right]^{-1} $$ $$ W'_{g}(x) = \left[r + \mu + \delta_{g} + \lambda_{gp} (1 - F_{p}(R_{gp}(w))) + \lambda_{gg} (1 - F_{g}(w))\right]^{-1} $$

The result should be: $$ R'_{pg}(w) = \frac{r + \mu + \delta_g + \lambda_{gp}(1 - F_{p}(w)) + \lambda_{gg}(1 - F_{g}(R_{pg}(w)))}{r + \mu + \delta_p + \lambda_{pp}(1 - F_{p}(w)) + \lambda_{pg}(1 - F_{g}(R_{pg}(w)))} $$

(I left a couple of terms out as they should not depend on w.)

Johan
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1 Answers1

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To get you started: By Fundamental Theorem of Calculus, $$ \frac{d}{dw}\int_{w}^{\infty} W'_{p}(x)(1 - F_{p}(x)) \, dx = -W'_p(w)(1-F_p(w)) $$

By the chain rule, $$ \frac{d}{dw}\int_{R_{pg}(w)}^{\infty} W'_{g}(x)(1 - F_{g}(x)) \, dx = -W'_g(R_{pg}(w))(1-F_g(R_{pg}(w)))R'_{pg}(w) $$

Note that after differentiating your $R_{pg}(w)$, the term $R'_{pg}(w)$ appears on both sides so you need to do some manipulation.

Benjamin Wang
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  • Thank you so much! This helped me solve it. This was my first post here, and it is fantastic to get help with a problem like this so quickly! – Johan Oct 27 '23 at 07:39
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    Thanks. If it helped, please consider upvoting and clicking the "tick" mark to accept my answer. – Benjamin Wang Oct 27 '23 at 07:54
  • ok, I was not aloud to vote as I am new, but I ticked it. – Johan Oct 27 '23 at 08:47