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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

= \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

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

Simplify

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

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

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

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

f (x) = x^{2} - x + 7, at x = 2

x y - 5 y^{2} = 4 (12.2)

\int_{- 3}^{4} \frac{8}{x ^{3}} d x

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