(\partial)/(\partial α)({K}(α)(α)^αL^{1-α})

解

\frac{\partial}{\partial α} (K (α) (α)^{α} L^{1 - α})

\frac{\partial}{\partial α} (K (α)) α^{α} L^{1 - α} + (L^{1 - α} α^{α} (ln (α) + 1) - L^{1 - α} α^{α} ln (L)) K (α)

解答ステップ

\frac{\partial}{\partial α} (K (α) (α)^{α} L^{1 - α})

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = K (α), g = α^{α} L^{1 - α}

= \frac{\partial}{\partial α} (K (α)) α^{α} L^{1 - α} + \frac{\partial}{\partial α} (α^{α} L^{1 - α}) K (α)

\frac{\partial}{\partial α} (α^{α} L^{1 - α}) = L^{1 - α} α^{α} (ln (α) + 1) - L^{1 - α} α^{α} ln (L)

= \frac{\partial}{\partial α} (K (α)) α^{α} L^{1 - α} + (L^{1 - α} α^{α} (ln (α) + 1) - L^{1 - α} α^{α} ln (L)) K (α)