What is (\partial)/(\partial x)(-4y^2x(6-x^2y^2)) ?

The answer to (\partial)/(\partial x)(-4y^2x(6-x^2y^2)) is -4y^2(6-3y^2x^2)

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(\partial)/(\partial x)(-4y^2x(6-x^2y^2))

\frac{\partial}{\partial x} (- 4 y^{2} x (6 - x^{2} y^{2}))

- 4 y^{2} (6 - 3 y^{2} x^{2})

Solution steps

\frac{\partial}{\partial x} (- 4 y^{2} x (6 - x^{2} y^{2}))

Treat

y

as a constant

Take the constant out:

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

= - 4 y^{2} \frac{\partial}{\partial x} (x (6 - x^{2} y^{2}))

Apply the Product Rule:

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

f = x, g = 6 - x^{2} y^{2}

= - 4 y^{2} (\frac{\partial}{\partial x} (x) (6 - x^{2} y^{2}) + \frac{\partial}{\partial x} (6 - x^{2} y^{2}) x)

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

\frac{\partial}{\partial x} (6 - x^{2} y^{2}) = - 2 y^{2} x

= - 4 y^{2} (1 \cdot (6 - x^{2} y^{2}) + (- 2 y^{2} x) x)

Simplify

- 4 y^{2} (1 \cdot (6 - x^{2} y^{2}) + (- 2 y^{2} x) x) : - 4 y^{2} (6 - 3 y^{2} x^{2})

= - 4 y^{2} (6 - 3 y^{2} x^{2})

(x^{2} + 1) \cdot \frac{d y}{d x} + 3 x (y - 1) = 0

x \to 5 lim (3 x + 2)

y^{^{''}} - 6 y^{^{'}} - 7 y = 0

(- 40) (- 4 - 10)

6 x - 32 x

What is (\partial)/(\partial x)(-4y^2x(6-x^2y^2)) ?
The answer to (\partial)/(\partial x)(-4y^2x(6-x^2y^2)) is -4y^2(6-3y^2x^2)