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

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

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

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

x (- x - 1)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (2) = 0

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

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

= x (0 + 0 - x - 1)

Simplify

= x (- x - 1)

(1 - t) s L + (1 - L) - \frac{( 1 - L ) ^{2}}{2} + xg

(1 + t (x)^{2}) \frac{d ^{2}}{d t d x} (x) + x

- \frac{3}{4} (e^{- x} - e^{3 x})

\frac{V ( p + h ) ( V p - Vh ) ( h - p )}{p ^{2}}

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