What is (\partial)/(\partial x)((y)(8-4x^2-2z^2)) ?

The answer to (\partial)/(\partial x)((y)(8-4x^2-2z^2)) is -8yx

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(\partial)/(\partial x)((y)(8-4x^2-2z^2))

\frac{\partial}{\partial x} ((y) (8 - 4 x^{2} - 2 z^{2}))

- 8 y x

Solution steps

\frac{\partial}{\partial x} ((y) (8 - 4 x^{2} - 2 z^{2}))

Treat

y, z

as constants

Take the constant out:

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

= y \frac{\partial}{\partial x} (8 - 4 x^{2} - 2 z^{2})

Apply the Sum/Difference Rule:

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

= y (\frac{\partial}{\partial x} (8) - \frac{\partial}{\partial x} (4 x^{2}) - \frac{\partial}{\partial x} (2 z^{2}))

\frac{\partial}{\partial x} (8) = 0

\frac{\partial}{\partial x} (4 x^{2}) = 8 x

\frac{\partial}{\partial x} (2 z^{2}) = 0

= y (0 - 8 x - 0)

Simplify

= - 8 y x

(a + b F)^{2} \cdot F^{2} \cdot (1 - c F - d F^{2} - e F^{3})

2 cos (x) (cos (x) + 1)

4 (cos (x) - cos (2 x))

(4 t - t^{2}, t - (1 - t)^{2}, 1 - t - 16 t^{2}) \cdot (4, 1, - 1)

- \frac{4}{( 2 x + 1 ) ^{2}}

What is (\partial)/(\partial x)((y)(8-4x^2-2z^2)) ?
The answer to (\partial)/(\partial x)((y)(8-4x^2-2z^2)) is -8yx