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

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

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

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

- \frac{3}{z}

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

= \frac{1}{z} (0 - 3)

Simplify

\frac{1}{z} (0 - 3) : - \frac{3}{z}

= - \frac{3}{z}

f (x) = x^{2} + 8, (- 3, 17)

\int_{2}^{7} \frac{1}{x ^{2} - 1} d x

y = \frac{6 x}{x + 2}

\frac{d}{d x} ((1 + 5 x) (1 - 7 x))

\frac{d}{d x} ((lo g_{e} (x))^{x})

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