What is (\partial)/(\partial L)({K}(α)(α)^αL^{1-α}) ?

The answer to (\partial)/(\partial L)({K}(α)(α)^αL^{1-α}) is L^{-α}α^α{K}(α)(1-α)

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

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

L^{- α} α^{α} K (α) (1 - α)

Solution steps

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

Treat

α

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= K (α) α^{α} (1 - α) L^{1 - α - 1}

Simplify

= L^{- α} α^{α} K (α) (1 - α)

y^{^{'}} + 4 y = 0

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\int 18 x^{2} (6 x^{3} + 5)^{\frac{1}{2}} d x

\int tan^{2} (x) (sec^{2} (x) - 1) d x

\frac{d}{d x} ((x^{3} + x)^{\frac{3}{2}})

What is (\partial)/(\partial L)({K}(α)(α)^αL^{1-α}) ?
The answer to (\partial)/(\partial L)({K}(α)(α)^αL^{1-α}) is L^{-α}α^α{K}(α)(1-α)