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

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

Popular >

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\int_{0}^{1} π (3 x)^{2} d x

\frac{d}{d x} (7^{x})

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 4 + e^{\frac{x}{2}}

n = 1 \sum \infty \frac{( 4 ^{n} + 1 )}{5 ^{n}}

\int - x e^{- 2 x} d x

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