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

The answer to (\partial)/(\partial x)(3/(4y^4x^{1/4)}) is -3/(16y^4x^{5/4)}

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

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

- \frac{3}{16 y ^{4} x ^{\frac{5}{4}}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

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

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

Simplify

\frac{3}{4 y ^{4}} (- \frac{1}{4} x^{- \frac{1}{4} - 1}) : - \frac{3}{16 y ^{4} x ^{\frac{5}{4}}}

= - \frac{3}{16 y ^{4} x ^{\frac{5}{4}}}

\frac{d x}{d t} = 7 x t^{4}

\int_{- 1}^{3} (2 x^{2} - 8) d x

\frac{d}{d x} (- \frac{18}{a x})

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

\int_{- 2}^{18} \frac{1}{∣ 2 x ∣} d x

What is (\partial)/(\partial x)(3/(4y^4x^{1/4)}) ?
The answer to (\partial)/(\partial x)(3/(4y^4x^{1/4)}) is -3/(16y^4x^{5/4)}