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Starting on p. 98 of Salanie's "The Economic of Taxation" (2nd edition), it explains

To probe it more rigorously, let us define the utility of taxpayer $w$ when he claims to have productivity $w'$: $$ V(w', w) = u(C(w'), Y(w'), w). $$ For the mechanism to be revealing, $V$ must be maximal in $w' = w$. Assume that all functions are differentiable and that income $Y$ is positive. Then we have the first-order necessary condition $$ \frac{\partial V}{\partial w'} (w, w) = 0 \tag{NC1} $$ and the second-order necessary condition $$ \frac{\partial^2 V}{\partial w'^2}(w,w) \leq 0 \tag{NC2}. $$ Differentiating (NC1) gives us $$ \frac{\partial^2 V}{\partial w'^2} + \frac{\partial^2 V}{\partial w' \partial w} = 0 $$ ...

My question is this: How do we arrive at the last equation from (NC1)? If I differentiate $$ \frac{\partial V}{\partial w'} (w', w) $$ with respect to $w$, don't I just end up with $\frac{\partial^2 V}{\partial w'\partial w}$? Likewise, if I differentiate with respect to $w'$, I end up with $\frac{\partial^2 V}{\partial w'^2}$. What am I missing?

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up vote 3 down vote accepted

I think it is the chain rule. Let $w'(w) = w$, since we are looking for revealing mechanisms. The condition $$ \frac{\partial V}{\partial w'} (w'(w),w) = 0 $$ holds for all $w$ because the mechanism is revealing for all types. As the (not partial) differentiate w.r.t. $w$ of the right hand side is 0, the same goes for the differentiate of the left hand size, hence $$ \frac{d \frac{\partial V}{\partial w'} (w'(w),w)}{d w} = 0. $$ According to the chain rule $$ \frac{d \frac{\partial V}{\partial w'} (w'(w),w)}{d w} = \frac{d w'(w)}{d w}\cdot \frac{\partial^2 V}{\partial^2 w'} (w'(w),w) + \frac{d w}{d w}\cdot \frac{\partial^2 V}{\partial w \partial w'} (w'(w),w), $$ which in a simplified form is $$ \frac{\partial^2 V}{\partial^2 w'} (w'(w),w) + \frac{\partial^2 V}{\partial w \partial w'} (w'(w),w). $$

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Why are you doing $\frac{\partial^2 V}{\partial w'^2}$ ?

Even if it is said that $w^{'}=w$ at the optimum, it should be taken different when you differentiate it for first order conditions. So, you differentiate it according to $w^{'}$ and $w$.

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