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

The answer to (\partial}{\partial y}(\frac{(x+y))/z) is 1/z

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

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

\frac{1}{z}

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (x) = 0

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

= \frac{1}{z} (0 + 1)

Simplify

= \frac{1}{z}

\frac{d}{d x} (\frac{2}{3} (- 2 e^{- 2 x} + e^{x}))

\int (sin (y)) d y

\int cos^{5} (x t a) n^{3} x d x

\frac{d}{d x} (\frac{2}{3 x})

a re a y = x, y = 1 - x, y = - \frac{x}{2}

What is (\partial}{\partial y}(\frac{(x+y))/z) ?
The answer to (\partial}{\partial y}(\frac{(x+y))/z) is 1/z