What is (\partial)/(\partial x)(-3cos(2x)) ?

The answer to (\partial)/(\partial x)(-3cos(2x)) is 6sin(2x)

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(\partial)/(\partial x)(-3cos(2x))

\frac{\partial}{\partial x} (- 3 cos (2 x))

6 sin (2 x)

Solution steps

\frac{\partial}{\partial x} (- 3 cos (2 x))

Take the constant out:

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

= - 3 \frac{\partial}{\partial x} (cos (2 x))

Apply the chain rule:

- sin (2 x) \frac{\partial}{\partial x} (2 x)

= - sin (2 x) \frac{\partial}{\partial x} (2 x)

\frac{\partial}{\partial x} (2 x) = 2

= - 3 (- sin (2 x) \cdot 2)

Simplify

= 6 sin (2 x)

\int \frac{5}{1 - 9 x ^{2}} + 7 x d x

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

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

f (x) = \frac{5}{x}, at x = 4, x = 5

y = (9 x^{2} - 18 x + 18) e^{- 4 x}

What is (\partial)/(\partial x)(-3cos(2x)) ?
The answer to (\partial)/(\partial x)(-3cos(2x)) is 6sin(2x)