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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derivative of

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)

y^{^{'}} + \frac{2 x}{1 + x ^{2}} y = \frac{e ^{- x}}{1 + x ^{2}}

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

x \to \infty lim (x^{- 1})

x^{\frac{4}{5}}

(3 x y^{2} + 13 y) + (3 x^{2} y + 13 x) y^{^{'}} = 0

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)