What is (\partial)/(\partial y)(x^ay^{1-a}) ?

The answer to (\partial)/(\partial y)(x^ay^{1-a}) is x^ay^{-a}(1-a)

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(\partial)/(\partial y)(x^ay^{1-a})

\frac{\partial}{\partial y} (x^{a} y^{1 - a})

x^{a} y^{- a} (1 - a)

Solution steps

\frac{\partial}{\partial y} (x^{a} y^{1 - a})

Treat

a, x

as a constant

Take the constant out:

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

= x^{a} \frac{\partial}{\partial y} (y^{1 - a})

Apply the Power Rule:

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

= x^{a} (1 - a) y^{1 - a - 1}

Simplify

= x^{a} y^{- a} (1 - a)

\frac{d}{d y} (y + 2 + y \frac{y + 2}{2} - \frac{( y + 2 ) ^{2}}{4})

x \to - 4 lim (x^{2} - 16)

\frac{d}{d x} (2 e)

\frac{d}{d x} (6 sin (x) - \frac{2}{cos ( x )})

e^{e^{x}} \to 0 lim (\frac{( e ^{x} ) ^{- 2}}{e ^{x}})

What is (\partial)/(\partial y)(x^ay^{1-a}) ?
The answer to (\partial)/(\partial y)(x^ay^{1-a}) is x^ay^{-a}(1-a)