的導數X小號1(p1,p2,X¯¯¯1,X¯¯¯2)≡X1(p1,p2,p1X¯¯¯1+p2X¯¯¯2)x1S(p1,p2,x¯1,x¯2)≡x1(p1,p2,p1x¯1+p2x¯2)x_1^S(p_1, p_2, overline{x}_…
為什麼偏導數 $ x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2) \equiv x_1(p_1,p_2,p_1\overline{x}_1+p_2\overline{x}_2) $ 為了 $ p_1 $ $$ \frac{\partial x_1^S(p_1, p_2, \overline{x}_1, \overline{x}_2)}{\partial p_1}=\frac{\partial x_1(p_1,p_2,\overline{m})}{\partial p_1} +\frac{\partial x_1(p_1,p_2,\overline{m})}{\partial m}\overline{x}_1 $$
在哪裡 $ (\overline{x}_1,\overline{x}_2) $ 是最初按價格需求的捆綁 $ (\overline{p}_1,\overline{p}_2) $ 和收入 $ \overline{m} $ 和 $ x_1^S $ 是商品 1 的斯盧茨基需求函式，並且 $ x_1 $ 商品 1 的馬歇爾需求函式？
我明白那個 $ m=p_1\overline{x}_1+p_2\overline{x}_2 $ 因為樞軸確保消費者仍然有足夠的收入來購買他的舊捆綁包 $ (\overline{x}_1,\overline{x}_2) $ .
Varian 的第 8 章展示了這一點，以使用微積分推導出 Slutsky 方程。我不明白等式的右邊。
我知道 $ p_1\overline{x}_1+p_2\overline{x}_2=m $ 和 $ \overline{p}_1\overline{x}_1+\overline{p}_2\overline{x}_2=\overline{m} $ 在哪裡 $ m $ 是新的預算。
This is more of a calculus question. Recall the total differential of a function $ f(z_1,z_2,z_3) $ :
$$ \begin{equation} \mathrm df(z_1,z_2,z_3)=\frac{\partial f(z_1,z_2,z_3)}{\partial z_1}\mathrm dz_1 + \frac{\partial f(z_1,z_2,z_3)}{\partial z_2}\mathrm dz_2+\frac{\partial f(z_1,z_2,z_3)}{\partial z_3}\mathrm dz_3. \end{equation} $$ Hence, $$ \begin{equation} \frac{\mathrm df(z_1,z_2,z_3)}{\mathrm dz_1}=\frac{\partial f(z_1,z_2,z_3)}{\partial z_1}\frac{\mathrm dz_1}{\mathrm dz_1} + \frac{\partial f(z_1,z_2,z_3)}{\partial z_2}\frac{\mathrm dz_2}{\mathrm dz_1} + \frac{\partial f(z_1,z_2,z_3)}{\partial z_3}\frac{\mathrm dz_3}{\mathrm dz_1} \end{equation} $$ Applying to your case, we substitute $ p_1 $ for $ z_1 $ , $ p_2 $ for $ z_2 $ , and $ \overline m $ for $ z_3 $ (noting that $ \overline m(p_1,p_2) $ is a function of the prices):
$$ \begin{align}\require{cancel} \frac{\mathrm dx_1(p_1,p_2,\overline m)}{\mathrm dp_1}&=\frac{\partial x_1(p_1,p_2,\overline m)}{\partial p_1}\cancelto{1}{\frac{\mathrm dp_1}{\mathrm dp_1}} + \frac{\partial x_1(p_1,p_2,\overline m)}{\partial p_2}\cancelto{0}{\frac{\mathrm dp_2}{\mathrm dp_1}} + \frac{\partial x_1(p_1,p_2,\overline m)}{\partial \overline m}\frac{\mathrm d\overline m}{\mathrm dp_1}\ &=\frac{\partial x_1(p_1,p_2,\overline m)}{\partial p_1} + \frac{\partial x_1(p_1,p_2,\overline m)}{\partial \overline m}\overline x_1 \end{align} $$