What is (\partial)/(\partial u)((x)(-3u-3v^2)) ?

The answer to (\partial)/(\partial u)((x)(-3u-3v^2)) is -3x

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(\partial)/(\partial u)((x)(-3u-3v^2))

\frac{\partial}{\partial u} ((x) (- 3 u - 3 v^{2}))

- 3 x

Solution steps

\frac{\partial}{\partial u} ((x) (- 3 u - 3 v^{2}))

Treat

x, v

as constants

Take the constant out:

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

= x \frac{\partial}{\partial u} (- 3 u - 3 v^{2})

Apply the Sum/Difference Rule:

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

= x (- \frac{\partial}{\partial u} (3 u) - \frac{\partial}{\partial u} (3 v^{2}))

\frac{\partial}{\partial u} (3 u) = 3

\frac{\partial}{\partial u} (3 v^{2}) = 0

= x (- 3 - 0)

Simplify

= - 3 x

cos (2 x - 4) + T g (5 x^{3} - 8)

e^{4 x} (4 sin (x) + cos (x))

A (A^{^{'}} + B + C)

\frac{csc ( x ) ( sec ( x ) - 1 ) - tan ( x )}{sec ( x ) - 1}

4 (3 + cos (2 x))^{2}

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