What is the derivative of 3e^{3x}cos(2x) ?

The derivative of 3e^{3x}cos(2x) is 3(3e^{3x}cos(2x)-2e^{3x}sin(2x))

What is the first derivative of 3e^{3x}cos(2x) ?

The first derivative of 3e^{3x}cos(2x) is 3(3e^{3x}cos(2x)-2e^{3x}sin(2x))

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\frac{d}{d x} (3 e^{3 x} cos (2 x))

3 (3 e^{3 x} cos (2 x) - 2 e^{3 x} sin (2 x))

Solution steps

\frac{d}{d x} (3 e^{3 x} cos (2 x))

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{3 x}, g = cos (2 x)

= 3 (\frac{d}{d x} (e^{3 x}) cos (2 x) + \frac{d}{d x} (cos (2 x)) e^{3 x})

\frac{d}{d x} (e^{3 x}) = e^{3 x} \cdot 3

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

= 3 (e^{3 x} \cdot 3 cos (2 x) + (- sin (2 x) \cdot 2) e^{3 x})

Simplify

= 3 (3 e^{3 x} cos (2 x) - 2 e^{3 x} sin (2 x))

t \frac{d y}{d t} + 3 y = t^{4}

\int_{0}^{3} (4 - x^{2}) d x

x \to 3 lim (4 x + 1)

f (x) = \frac{3 x}{e ^{x}}

f (x) = 2 e^{x} + x, (0, 2)

What is the derivative of 3e^{3x}cos(2x) ?
The derivative of 3e^{3x}cos(2x) is 3(3e^{3x}cos(2x)-2e^{3x}sin(2x))
What is the first derivative of 3e^{3x}cos(2x) ?
The first derivative of 3e^{3x}cos(2x) is 3(3e^{3x}cos(2x)-2e^{3x}sin(2x))