Answer to Question #230959 in Macroeconomics for workiye Misgana Ha

Question

1.    Given y = (x, z) = 6x² + 3xz⁴ where

X(B, E, T) = B² + 2E + TB

Z(B, E, T) = (ETB)¹ + B + E Then, find the partial derivatives of y with respect to B, E, and T.

Answer 1

Given that,

$y = (x, z) = 6x^2 + 3xz^4 \\where\\ X(B, E, T) = B^2 + 2E + TB\\ Z(B, E, T) = (ETB)1 + B + E \\ So,\\ y=6(B^2+2E+TB)^2+3(B^2+2E+TB)(ETB+B+E)^4$

Integrating partially with respect to B,

$\frac{∂y}{∂B}=\frac{∂}{∂B}[6(B^2+2E+TB)^2]+\frac{∂}{∂B}[6(B^2+2E+TB)(ETB+B+E)^4]\\=12(B^2+2E+TB)(2B+T)+3(2B+T)(ETB+B+E)^4+12(B^2+2E+TB)(ETB+B+E)^3(ET+1)$

Integrating partially with respect to E,

$\frac{∂y}{∂E}=\frac{∂}{∂E}[6(B^2+2E+TB)^2]+\frac{∂}{∂E}[6(B^2+2E+TB)(ETB+B+E)^4]\\=12(B^2+2E+TB)×2+3(ETB+B+E)^4×2+3(B^2+2E+TB)×4(ETB+B+E)^4(TB+1)\\=24(B^2+2E+TB)+6(ETB+B+E)^4+12(B^2+2E+TB)(ETB+B+E)^4(TB+1)$

Integrating partially with respect to T,

$\frac{∂y}{∂T}=12(B^2+2E+TB)B+3(ETB+B+E)^4B+12(B^2+2E+TB)(ETB+B+E)^3EB$