What is (\partial)/(\partial z)((z^2+2x+y)/(xy)) ?

The answer to (\partial)/(\partial z)((z^2+2x+y)/(xy)) is (2z)/(xy)

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(\partial)/(\partial z)((z^2+2x+y)/(xy))

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

\frac{2 z}{x y}

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

= \frac{1}{x y} \frac{\partial}{\partial z} (z^{2} + 2 x + y)

Apply the Sum/Difference Rule:

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

= \frac{1}{x y} (\frac{\partial}{\partial z} (z^{2}) + \frac{\partial}{\partial z} (2 x) + \frac{\partial}{\partial z} (y))

\frac{\partial}{\partial z} (z^{2}) = 2 z

\frac{\partial}{\partial z} (2 x) = 0

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

= \frac{1}{x y} (2 z + 0 + 0)

Simplify

\frac{1}{x y} (2 z + 0 + 0) : \frac{2 z}{x y}

= \frac{2 z}{x y}

\int p (p + 9)^{8} d p

\int \frac{2 x - 3}{x} d x

x + y \frac{d y}{d x} = 0, y (16) = 4

x = \frac{1}{y ^{2}}, x = 1, x = 4

\int \frac{- 8}{x ^{2} + 16} d x

What is (\partial)/(\partial z)((z^2+2x+y)/(xy)) ?
The answer to (\partial)/(\partial z)((z^2+2x+y)/(xy)) is (2z)/(xy)