What is (\partial)/(\partial x)(ax^by^{1-b}) ?

The answer to (\partial)/(\partial x)(ax^by^{1-b}) is ay^{1-b}bx^{b-1}

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(\partial)/(\partial x)(ax^by^{1-b})

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

a y^{1 - b} b x^{b - 1}

Solution steps

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

Treat

a, b, y

as constants

Take the constant out:

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

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

Apply the Power Rule:

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

= a y^{1 - b} b x^{b - 1}

\frac{d y}{d x} = x^{2} e^{x^{3}}

\frac{\partial}{\partial x} (4 sin (3 x))

x (x^{6} + 6)

\frac{\partial}{\partial x} (y^{3} cos (6 x))

y^{^{''}} - y^{^{'}} = sin (x)

What is (\partial)/(\partial x)(ax^by^{1-b}) ?
The answer to (\partial)/(\partial x)(ax^by^{1-b}) is ay^{1-b}bx^{b-1}