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

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

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

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

- \frac{1}{x + z}

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

Simplify

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

\frac{d y}{d x} = 7 x y

x \to 0 lim (3 x + 4)

\int \frac{cos ^{3} ( x ) + 4}{cos ^{2} ( x )} d x

\int 3 sin^{2} (x) cos^{2} (x) d x

\frac{\partial}{\partial y} (\frac{- x - 7 y}{y ^{2} + 2 x ^{2}})

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