What is (\partial)/(\partial x)(-2ycos(6x)) ?

The answer to (\partial)/(\partial x)(-2ycos(6x)) is 12ysin(6x)

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(\partial)/(\partial x)(-2ycos(6x))

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

12 y sin (6 x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

- sin (6 x) \frac{\partial}{\partial x} (6 x)

= - sin (6 x) \frac{\partial}{\partial x} (6 x)

\frac{\partial}{\partial x} (6 x) = 6

= - 2 y (- sin (6 x) \cdot 6)

Simplify

= 12 y sin (6 x)

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

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{6}})

\frac{d}{d x} (\frac{2}{1 - 2 x})

\frac{\partial}{\partial x} (\frac{x}{2 x ^{2} + y ^{2}})

\int x^{3} \cdot e^{12 x} d x

What is (\partial)/(\partial x)(-2ycos(6x)) ?
The answer to (\partial)/(\partial x)(-2ycos(6x)) is 12ysin(6x)