What is (\partial)/(\partial y)((x-y)/(xy+3)) ?

The answer to (\partial)/(\partial y)((x-y)/(xy+3)) is (-x^2-3)/((xy+3)^2)

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(\partial)/(\partial y)((x-y)/(xy+3))

\frac{\partial}{\partial y} (\frac{x - y}{x y + 3})

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

Solution steps

\frac{\partial}{\partial y} (\frac{x - y}{x y + 3})

Treat

x

as a constant

Apply the Quotient Rule:

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

= \frac{\frac{\partial}{\partial y} ( x - y ) ( x y + 3 ) - \frac{\partial}{\partial y} ( x y + 3 ) ( x - y )}{( x y + 3 ) ^{2}}

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

\frac{\partial}{\partial y} (x y + 3) = x

= \frac{( - 1 ) ( x y + 3 ) - x ( x - y )}{( x y + 3 ) ^{2}}

Simplify

\frac{( - 1 ) ( x y + 3 ) - x ( x - y )}{( x y + 3 ) ^{2}} : \frac{- x ^{2} - 3}{( x y + 3 ) ^{2}}

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

x \to 0 lim (x ln^{3} (1 + 3 x^{3}))

x^{4} + 3 e^{x}

x^{2} - 2 x

f (x) = - sech (x) tanh (x)

\int \frac{1}{u ( 1 + u ^{2} )} d u

What is (\partial)/(\partial y)((x-y)/(xy+3)) ?
The answer to (\partial)/(\partial y)((x-y)/(xy+3)) is (-x^2-3)/((xy+3)^2)