What is (\partial)/(\partial z)(2xyz^{-1}) ?

The answer to (\partial)/(\partial z)(2xyz^{-1}) is -(2xy)/(z^2)

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(\partial)/(\partial z)(2xyz^{-1})

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

- \frac{2 x y}{z ^{2}}

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

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

Apply the Power Rule:

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

= 2 x y (- 1 \cdot z^{- 1 - 1})

Simplify

2 x y (- 1 \cdot z^{- 1 - 1}) : - \frac{2 x y}{z ^{2}}

= - \frac{2 x y}{z ^{2}}

\int \frac{3 x ^{2}}{1 - x ^{3}} d x

y^{2} d x + (2 x y + 1) d y = 0, y (9) = 90

x \to 0 + lim (tan^{x} (2 x))

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

\int e^{- x} \cdot ln (x) d x

What is (\partial)/(\partial z)(2xyz^{-1}) ?
The answer to (\partial)/(\partial z)(2xyz^{-1}) is -(2xy)/(z^2)