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

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

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

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

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

Solution steps

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

Treat

a, y, b

as constants

Take the constant out:

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

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

Apply the Power Rule:

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

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

\frac{\partial}{\partial x} (\frac{( 2 x )}{y})

y^{^{''}} - 4 y = 6 e^{t}

5 cos (2 x), 5 - 5 cos (2 x), 0, \frac{π}{2}

\int \frac{20 x ^{3}}{4 + 5 x ^{4}} d x

y^{^{''}} + 4 y = t \cdot sin (2 t)

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