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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

\int \frac{e ^{x}}{1 + 25 e ^{2 x}} d x

f (x) = 2 x + \frac{3}{x}, at x = - 1

\frac{\partial}{\partial x} (\frac{arctan ( \cdot y ^{2} )}{x})

\frac{d y}{d t} + e^{t} y = - e^{t}

2 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

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