What is (\partial)/(\partial x)((z-x)/(z+y)) ?

The answer to (\partial)/(\partial x)((z-x)/(z+y)) is -1/(z+y)

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(\partial)/(\partial x)((z-x)/(z+y))

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

- \frac{1}{z + y}

Solution steps

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

Treat

z, y

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= - \frac{1}{z + y}

\frac{d}{d x} (\frac{16 x}{( 4 - x ^{2} ) ^{2}})

\int \frac{1}{2 x ^{2} + 3 x + 1} d x

\frac{\partial}{\partial x} (4 sin (6 x - 7 y))

\int - 2 t^{5} d t

(x^{2} - 9)^{2}

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