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

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

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(\partial)/(\partial x)(x^{3/4}y^{1/4})

\frac{\partial}{\partial x} (x^{\frac{3}{4}} y^{\frac{1}{4}})

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

Solution steps

\frac{\partial}{\partial x} (x^{\frac{3}{4}} y^{\frac{1}{4}})

Treat

y

as a constant

Take the constant out:

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

= y^{\frac{1}{4}} \frac{\partial}{\partial x} (x^{\frac{3}{4}})

Apply the Power Rule:

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

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

Simplify

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

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

x \to 0 - lim (\frac{1}{e ^{\frac{1}{x}} - 1})

(x^{3} - x) (x^{2} + 2) (x^{2} + x - 1)

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

\int (2 x^{2} + 1)^{2} d x

\frac{d}{d x} (x^{2} - 1)

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