What is (\partial)/(\partial x)(-4/(x^2y)) ?

The answer to (\partial)/(\partial x)(-4/(x^2y)) is 8/(x^3y)

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(\partial)/(\partial x)(-4/(x^2y))

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

\frac{8}{x ^{3} y}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

- \frac{4}{y} (- 2 x^{- 2 - 1}) : \frac{8}{x ^{3} y}

= \frac{8}{x ^{3} y}

x^{^{''}} - 20 x^{^{'}} + 91 x = 0

f (t) = t^{\frac{1}{2}} (13 t + 2)

\frac{9}{( 9 - x ^{8} ) ^{\frac{3}{2}}}

\int_{3}^{5} \frac{x ^{2}}{8} d x

n = 1 \sum \infty \frac{n}{1000 n + 1}

What is (\partial)/(\partial x)(-4/(x^2y)) ?
The answer to (\partial)/(\partial x)(-4/(x^2y)) is 8/(x^3y)