What is (\partial)/(\partial x)((x+y)/(xy-1)) ?

The answer to (\partial)/(\partial x)((x+y)/(xy-1)) is (-y^2-1)/((xy-1)^2)

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(\partial)/(\partial x)((x+y)/(xy-1))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = sin (y) + 1

\frac{d}{d x} (x^{3 c o s (x)})

f (x) = 7 x

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

What is (\partial)/(\partial x)((x+y)/(xy-1)) ?
The answer to (\partial)/(\partial x)((x+y)/(xy-1)) is (-y^2-1)/((xy-1)^2)