What is (\partial}{\partial z}(\frac{(x+y))/z) ?

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

= (x + y) \frac{\partial}{\partial z} (\frac{1}{z})

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

y = \frac{x ^{2} - 8 x + 3}{x ^{2} + 5}

\frac{d}{d x} (ln (x^{c o s (x)}))

x y + 1 + e = e^{- x y}, (1, - 1)

\int 6 x y^{2} d y

f (x) = e^{x}, at x = ln (14)

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