What is (\partial)/(\partial y)(x^ay^b) ?

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

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(\partial)/(\partial y)(x^ay^b)

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

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

Solution steps

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

Treat

a, x, b

as constants

Take the constant out:

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

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

Apply the Power Rule:

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

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

\int_{0}^{π} sec^{2} (x) d x

c y^{- 4}

\int (3 x^{\frac{1}{3}}) d x

\int (e^{3 x + 9} - 3 x^{2}) d x

y = 8 arcsin (\frac{x}{4}) - \frac{x 16 - x ^{2}}{2}

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