What is (\partial)/(\partial y)(6x^{1/3}y^{2/3}) ?

The answer to (\partial)/(\partial y)(6x^{1/3}y^{2/3}) is (4x^{1/3})/(y^{1/3)}

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(\partial)/(\partial y)(6x^{1/3}y^{2/3})

\frac{\partial}{\partial y} (6 x^{\frac{1}{3}} y^{\frac{2}{3}})

\frac{4 x ^{\frac{1}{3}}}{y ^{\frac{1}{3}}}

Solution steps

\frac{\partial}{\partial y} (6 x^{\frac{1}{3}} y^{\frac{2}{3}})

Treat

x

as a constant

Take the constant out:

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

= 6 x^{\frac{1}{3}} \frac{\partial}{\partial y} (y^{\frac{2}{3}})

Apply the Power Rule:

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

= 6 x^{\frac{1}{3}} \frac{2}{3} y^{\frac{2}{3} - 1}

Simplify

6 x^{\frac{1}{3}} \frac{2}{3} y^{\frac{2}{3} - 1} : \frac{4 x ^{\frac{1}{3}}}{y ^{\frac{1}{3}}}

= \frac{4 x ^{\frac{1}{3}}}{y ^{\frac{1}{3}}}

\frac{\partial}{\partial x} (\frac{4 x}{( y + 2 z ) ^{3}})

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\frac{d}{d x} (\frac{2 x}{x - 3})

f (x) = \frac{3 + x}{1 - 3 x}

What is (\partial)/(\partial y)(6x^{1/3}y^{2/3}) ?
The answer to (\partial)/(\partial y)(6x^{1/3}y^{2/3}) is (4x^{1/3})/(y^{1/3)}