What is (\partial}{\partial y}(\frac{(3x+y))/z) ?

The answer to (\partial}{\partial y}(\frac{(3x+y))/z) is 1/z

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(\partial}{\partial y}(\frac{(3x+y))/z)

\frac{\partial}{\partial y} (\frac{( 3 x + y )}{z})

\frac{1}{z}

Solution steps

\frac{\partial}{\partial y} (\frac{( 3 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} (3 x + y)

Apply the Sum/Difference Rule:

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

= \frac{1}{z} (\frac{\partial}{\partial y} (3 x) + \frac{\partial}{\partial y} (y))

\frac{\partial}{\partial y} (3 x) = 0

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

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

Simplify

= \frac{1}{z}

\frac{d}{d x} (4^{5 x})

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\frac{d}{d x} ((\frac{6}{7})^{\frac{2}{3}})

\int x csc (\frac{π - x ^{2}}{3}) d x

f (x) = 4 arcsin (x^{4})

What is (\partial}{\partial y}(\frac{(3x+y))/z) ?
The answer to (\partial}{\partial y}(\frac{(3x+y))/z) is 1/z