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

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

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

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

\frac{2 x}{z}

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the common derivative:

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

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

Simplify

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

= \frac{2 x}{z}

y^{^{'}} = e^{5 y + 8 x}

\int \frac{1}{e ^{3 x}} d x

y d x - x d y + x y d y = 0

n = 0 \sum \infty \frac{2}{n ( n + 2 )}

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

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