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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 2 \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}

= 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 y^{2} x

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

Simplify

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

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

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\int arctan (2 t) d t

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