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

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

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

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

2 x y^{2} - 3 x^{2} + 2 y

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

f = x^{2} - 2 y, g = y^{2} - x

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

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

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

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

Simplify

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

= 2 x y^{2} - 3 x^{2} + 2 y

\frac{d}{d x} (\frac{x + π}{x} cos (x))

\int 2^{7 - x} d x

\int \frac{e ^{- x}}{cos ^{2} ( e ^{- x} )} d x

ln (x^{2} - 7)

\int \frac{x - 1}{1 - x ^{2}} d x

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