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

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

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

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

- y^{2} + 2 x y - 1

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x - y, g = x y - 1

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

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

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

= 1 \cdot (x y - 1) + y (x - y)

Simplify

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

= - y^{2} + 2 x y - 1

\frac{d}{d x} (\frac{x \cdot sin ( x )}{1 + cos ( x )})

\int (x - 3) (x + 3) d x

\frac{d y}{d x} - \frac{1 y}{x} = 1 - \frac{1}{x}

e^{- 3 x} - 3 e^{- 3 x} x

\int 12 x ln (7 x) d x

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