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}}

\frac{\partial}{\partial x} (x y + y)

\frac{d}{d x} (x^{3} y^{3})

\frac{\partial}{\partial x} (3 x e^{x y})

x sin (y)

x \to - \infty lim (- 2 x - 1)

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