What is the derivative of-2cos(3x) ?

The derivative of-2cos(3x) is 6sin(3x)

What is the first derivative of-2cos(3x) ?

The first derivative of-2cos(3x) is 6sin(3x)

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

6 sin (3 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

- sin (3 x) \frac{d}{d x} (3 x)

= - sin (3 x) \frac{d}{d x} (3 x)

\frac{d}{d x} (3 x) = 3

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

Simplify

= 6 sin (3 x)

x \to 0 + lim (\frac{x}{3 x} - 7 x)

\int \frac{3}{91} x^{2} d x

\frac{\partial}{\partial x} (\frac{x + y}{x + z})

\frac{d x}{d t} = - \frac{x}{10} + \frac{1}{10 e ^{\frac{t}{10}}}

t y^{^{'}} - y = 3 t^{2} cos (2 t)

What is the derivative of-2cos(3x) ?
The derivative of-2cos(3x) is 6sin(3x)
What is the first derivative of-2cos(3x) ?
The first derivative of-2cos(3x) is 6sin(3x)