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

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

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

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

- 2 y x

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (1) = 0

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

= y (0 - 2 x)

Simplify

= - 2 y x

x \to 2 + lim (9 - x^{2})

\int_{1}^{2} ln (3) x^{2} d x

\frac{d y}{d x} = \frac{x - 4}{y + 3}

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

\frac{d}{d x} (sin (x) \cdot ln (x))

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