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

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

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(\partial)/(\partial y)((x^2-y^2)/(xy))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

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

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

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

Simplify

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

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

\int \frac{2 x}{x ^{2} + 5} d x

f (x) = x - x^{3}

\int ln (y) + 4 x y^{3} d x

\frac{d}{d x} (4 x - x^{2} y + y^{3})

\frac{s + 3}{s ^{2} + 4 s + 5}

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