What is (\partial)/(\partial x)((-y^4)/(4x^4)) ?

The answer to (\partial)/(\partial x)((-y^4)/(4x^4)) is (y^4)/(x^5)

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(\partial)/(\partial x)((-y^4)/(4x^4))

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

\frac{y ^{4}}{x ^{5}}

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

= \frac{y ^{4}}{x ^{5}}

\int (\frac{1}{1 - x}) d x

f (x) = 2 x^{2} + 2 x, (- 2, - 6)

\int_{0}^{2} ln (1 + x^{4}) d x

\int 3^{- x} d x

x^{^{''}} = - x + 3 x^{^{'}}

What is (\partial)/(\partial x)((-y^4)/(4x^4)) ?
The answer to (\partial)/(\partial x)((-y^4)/(4x^4)) is (y^4)/(x^5)