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

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

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

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

\frac{- z ^{2} - 2 x}{x y ^{2}}

Solution steps

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

Treat

z, x

as constants

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{1}{x} \cdot \frac{1 \cdot y - 1 \cdot ( z ^{2} + 2 x + y )}{y ^{2}} : \frac{- z ^{2} - 2 x}{x y ^{2}}

= \frac{- z ^{2} - 2 x}{x y ^{2}}

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

\int \frac{( y - 2 ) ^{2}}{y} d y

- 5 t^{4}

y = arctanh (\frac{1}{x})

\frac{d}{d x} (\frac{t}{x})

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