What is (\partial)/(\partial x)(x^{1/4}*y^{1/2}) ?

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

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(\partial)/(\partial x)(x^{1/4}*y^{1/2})

\frac{\partial}{\partial x} (x^{\frac{1}{4}} \cdot y^{\frac{1}{2}})

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

Solution steps

\frac{\partial}{\partial x} (x^{\frac{1}{4}} y^{\frac{1}{2}})

Treat

y

as a constant

Take the constant out:

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

= y^{\frac{1}{2}} \frac{\partial}{\partial x} (x^{\frac{1}{4}})

Apply the Power Rule:

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

= y^{\frac{1}{2}} \frac{1}{4} x^{\frac{1}{4} - 1}

Simplify

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

= \frac{y ^{\frac{1}{2}}}{4 x ^{\frac{3}{4}}}

(3 x^{2} + 3 x^{- 2} y) d x + (x^{2} y - x^{- 1}) d y = 0

\int_{0}^{2} π (x^{2} - L)^{2} d x

\frac{2 x + 1}{x + 3}

y = \frac{1}{x}, y = 4 x, y = \frac{1}{4} x

f (t) = t cos (t) - sin (t)

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