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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

n = 0 \sum \infty (- 1)^{n}

\int (cos (x) + 5^{x}) d x

\int (3^{x}) d x

y^{^{'}} = \frac{x y + 3 x - y - 3}{x y - 2 x + 4 y - 8}

y = \frac{x ^{2} + 5}{x - 1}

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