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

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

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

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

- \frac{2 y}{z}

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the common derivative:

\frac{\partial}{\partial x} (x) = 1

= - 2 y z^{- 1} \cdot 1

Simplify

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

= - \frac{2 y}{z}

x \to - 5 lim (\frac{x}{x + 4})

n = 0 \sum \infty n \cdot sin (nπ)

x \to 6 lim (\frac{∣ x - 6 ∣}{x - 6})

\int_{0}^{1} 6 (1 + x)^{7} d x

f (x) = (2 - x^{3})^{2}

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