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I am preparing myself for a MSc in Economics and don't understand why, according to the Envelope Theorem, when deriving a utility function $ u(θ, q(θ), r(θ)) = B(q-r) - C(\frac{q}{θ})$, its derivative with respect to $θ$ is $(\frac{q(θ)}{θ^2})C'(\frac{q(θ)}{θ})$.

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  • $\begingroup$ Would be helpful to explain what $B$ and $C$ functions are. $\endgroup$ – Art Jul 21 at 8:26
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I'm guessing your inquiry is concerning the chain rule of total derivatives.

In the special case that $B(q-r)$ is for some reason omitted out of the expression, the derivation sequence would be this.

$$\frac{\partial{u}}{{\partial\theta}}= \frac{\partial{u}}{{\partial{C}}}\frac{\partial{C}}{{\partial\theta}} = -1*\frac{\partial{C}}{{\partial\theta}}=-\frac{\partial{\frac{q(\theta)}{\theta}}}{{\partial\theta}}C′(\frac{q(θ)}{θ})=-(-\frac{q(θ)}{θ^2})C′(\frac{q(θ)}{θ})$$

If your problem is with the latter part of the expression above, I'd suggest you read through derivation rule of composite functions.

As @Giskard has pointed out, it is unclear from the question why the term $B(q-r)$ would be omitted and perhaps you'd like to provide additional information on that.

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  • $\begingroup$ Whilst I've come across the use of complex and even quaternion differentiation in physics. I've yet to see one in economics. Nor does the wiki page on the envelope theorem use complex variables - its given in terms of real variables. Do you have a reference for your assertion? $\endgroup$ – Mozibur Ullah Jul 8 at 12:33
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    $\begingroup$ I'm certain this comment is not targeted to my answer, but to the original question. I'm only providing the basic calculus method to derive the results madam Katia asked for. I don't think a person of your accumulated knowledge would have me back mathematics. In either case, the book we used on the subject is Fundamental methods of mathematical economics by Alpha Chiang and Kevin Wainwright. $\endgroup$ – the_rainbox Jul 8 at 12:40
  • $\begingroup$ Contrary to your assertion, I can affirm that my comment is targeted at your answer and not the question. $\endgroup$ – Mozibur Ullah Jul 8 at 12:43
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    $\begingroup$ @the_rainbox Downvoters are plentiful, don't worry about them. Also, IMO you should not have answered the question, as it is unclear. If $q$ is a function of $\theta$ why does $B(q-r)$ disappear? Parts seem to be missing. $\endgroup$ – Giskard Jul 8 at 12:54
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    $\begingroup$ @Mozibur Ullah The actual term would probably be "composite function". The translation from my native language came as "complex", sorry for the inconvenience. $\endgroup$ – the_rainbox Jul 8 at 13:51
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I think the answer above needs a little more clarification. The Envelope theorem is slightly more subtle than simple chain rule for derivatives. The idea is as follows: $\theta$ is a parameter that is held fixed in the optimization problem. So the optimal solution for $q,r$ depends on $\theta$ and is hence written as $q(\theta)$ and $r(\theta)$ and the function $u(\theta, q(\theta),r(\theta))$ is the indirect utility function (or value function). The envelope theorem states: around a small neighbourhood of the optimal solution, the derivative of the indirect utility function with respect to $\theta$ is the same as the derivative of the original utility function with respect to $\theta$ while holding $q,r$ constant (in other words, you can ignore that $q,r$ depends on $\theta$).

So lets do the exercise: Simple chain rule dictates that

$\frac{\partial u(\theta,q(\theta),r(\theta))}{\partial\theta} = q^{\prime}(\theta)\frac{\partial B(q(\theta)-r(\theta))}{\partial q}-r^{\prime}(\theta)\frac{\partial B(q(\theta)-r(\theta))}{\partial r}-\frac{\theta q^{\prime}(\theta)-q(\theta)}{\theta^2}\frac{\partial C(\frac{q}{\theta})}{\partial q}$.

Rearranging, we can write: $\frac{\partial u(\theta,q(\theta),r(\theta))}{\partial\theta} = -r^{\prime}(\theta)\big[\frac{\partial B(q(\theta)-r(\theta))}{\partial r}\big] + q^{\prime}(\theta)\big[\frac{\partial B(q(\theta)-r(\theta))}{\partial q} - \frac{1}{\theta}\frac{\partial C(\frac{q}{\theta})}{\partial q}\big] +\frac{q(\theta)}{\theta^2}\frac{\partial C(\frac{q(\theta)}{\theta})}{\partial q}$

At the optimum, $q(\theta),r(\theta)$ must satisfy the FOC (interior assumed):

$\frac{\partial B(q(\theta)-r(\theta))}{\partial r}=0$, and $\frac{\partial B(q(\theta)-r(\theta))}{\partial q} - \frac{1}{\theta}\frac{\partial C(\frac{q}{\theta})}{\partial q} = 0$

Plugging the FOCs into the the chain-rule expression then leaves us with our answer:

$\frac{\partial u(\theta,q(\theta),r(\theta))}{\partial\theta} = \frac{q(\theta)}{\theta^2}\frac{\partial C(\frac{q(\theta)}{\theta})}{\partial q}$

The last expresssion is exactly what you would get if you differentiated the original utility function $u(\theta,q,r)$ with respect to $\theta$ holding $q,r,$ fixed. $\bigg($ i.e. $\frac{\partial u(\theta, q,r)}{\partial\theta} =\frac{q}{\theta^2}\frac{\partial C(\frac{q}{\theta})}{\partial q} \bigg)$

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  • $\begingroup$ You don't actually need to show the whole derivation. It's a theorem that around a stationary point any perturbation can be more or less ignored. That's why mathematicians are interested in such points. $\endgroup$ – Mozibur Ullah Jul 8 at 17:29
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    $\begingroup$ I would rather be helpful with a simple explanation than obfuscate my answer with unnecessary jargon. It detracts from the point and is absolutely useless for someone who is willing to learn. And yet in all your obsession with jargon, you failed to address the prime issue of the answer by @the_rainbox: $B(q-r)$ being omitted is not a special case, but is a direct application of the theorem. So let me stick to my explanations, and Ill let you stick to your jargon. Thank you very much. $\endgroup$ – Tomcat Jul 8 at 17:44
  • $\begingroup$ That issue wasn't raised by the_rainbox, but by Giskard. If you check the comments you will see I'm asking for an apology from him for making 'careless' claims. A stationary point is hardly specialist jargon when it comes to calculus. It's terminology usually introduced in high school calculus. $\endgroup$ – Mozibur Ullah Jul 8 at 17:53
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    $\begingroup$ Ive checked the comments! My reaction stems from your initial comment here as well as in the other answer! Stating that the envelope theorem is a theorem around a stationary point is absolutely true, but it does not help someone who is trying to understand how the particular expression in the question is derived. Yes, giskard did raise the point, but the_rainbox agreed to it and modified the answer to incorporate the point. So my answer aims to clarify that. I cant really clarify it without showing the basic step in the calculation. $\endgroup$ – Tomcat Jul 8 at 18:00
  • $\begingroup$ *a theorem that allows you to ignore parameter perturbations around $\endgroup$ – Tomcat Jul 8 at 18:25

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