What is (\partial)/(\partial x)(xy(1-x^2-y^2)) ?

The answer to (\partial)/(\partial x)(xy(1-x^2-y^2)) is y(-3x^2-y^2+1)

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(\partial)/(\partial x)(xy(1-x^2-y^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = 1 - x^{2} - y^{2}

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{2}{x - 4})

\int \frac{x}{4 x + 1} d x

(6 x y - 2 y^{2} + 1) d x + (3 x^{2} - 4 x y) d y = 0

\int \frac{5}{6 x + 5} d x

\frac{x}{7 x ^{2} + 1}

What is (\partial)/(\partial x)(xy(1-x^2-y^2)) ?
The answer to (\partial)/(\partial x)(xy(1-x^2-y^2)) is y(-3x^2-y^2+1)