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

The derivative of-(3x+2cos(2x)) is -3cos(2x)+6xsin(2x)+4sin(2x)

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

The first derivative of-(3x+2cos(2x)) is -3cos(2x)+6xsin(2x)+4sin(2x)

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

- 3 cos (2 x) + 6 x sin (2 x) + 4 sin (2 x)

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = 3 x + 2, g = cos (2 x)

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

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

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

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

Simplify

- (3 cos (2 x) + (- sin (2 x) \cdot 2) (3 x + 2)) : - 3 cos (2 x) + 6 x sin (2 x) + 4 sin (2 x)

= - 3 cos (2 x) + 6 x sin (2 x) + 4 sin (2 x)

f (x) = sin (x) + \frac{1}{2} cot (x)

n = 0 \sum \infty \frac{n}{2 n + 4}

\int - \frac{sin ( x ) tan ( x )}{1} d x

\frac{d y}{d t} - \frac{4 t}{( 1 + 4 t ^{2} )} y = 4 t

y^{^{'}} = x^{2} sec (y)

What is the derivative of-(3x+2cos(2x)) ?
The derivative of-(3x+2cos(2x)) is -3cos(2x)+6xsin(2x)+4sin(2x)
What is the first derivative of-(3x+2cos(2x)) ?
The first derivative of-(3x+2cos(2x)) is -3cos(2x)+6xsin(2x)+4sin(2x)