What is (\partial}{\partial L}(L^{1/2)/K ^{1/2}) ?

The answer to (\partial}{\partial L}(L^{1/2)/K ^{1/2}) is ({K}^{1/2})/(2L^{1/2)}

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(\partial}{\partial L}(L^{1/2)/K ^{1/2})

\frac{\partial}{\partial L} (L^{\frac{1}{2}} K^{\frac{1}{2}})

\frac{K ^{\frac{1}{2}}}{2 L ^{\frac{1}{2}}}

Solution steps

\frac{\partial}{\partial L} (L^{\frac{1}{2}} K^{\frac{1}{2}})

Take the constant out:

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

= K^{\frac{1}{2}} \frac{\partial}{\partial L} (L^{\frac{1}{2}})

Apply the Power Rule:

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

= K^{\frac{1}{2}} \frac{1}{2} L^{\frac{1}{2} - 1}

Simplify

K^{\frac{1}{2}} \frac{1}{2} L^{\frac{1}{2} - 1} : \frac{K ^{\frac{1}{2}}}{2 L ^{\frac{1}{2}}}

= \frac{K ^{\frac{1}{2}}}{2 L ^{\frac{1}{2}}}

\frac{\partial}{\partial y} (x + cos (x yz) - 1)

2 x, x^{2 - 3}

\frac{d}{d x} (sin (x) + \frac{2}{9} cot (x))

ln (14 x)

\frac{d}{d x} (\frac{108}{x})

What is (\partial}{\partial L}(L^{1/2)/K ^{1/2}) ?
The answer to (\partial}{\partial L}(L^{1/2)/K ^{1/2}) is ({K}^{1/2})/(2L^{1/2)}