What is the derivative of f(x)=6cos(2x-pi/3)+3x ?

The derivative of f(x)=6cos(2x-pi/3)+3x is 6(-sin(2x)+sqrt(3)cos(2x))+3

What is the first derivative of f(x)=6cos(2x-pi/3)+3x ?

The first derivative of f(x)=6cos(2x-pi/3)+3x is 6(-sin(2x)+sqrt(3)cos(2x))+3

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derivative of

f (x) = 6 cos (2 x - \frac{π}{3}) + 3 x

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

Solution steps

\frac{d}{d x} (6 cos (2 x - \frac{π}{3}) + 3 x)

Simplify

6 cos (2 x - \frac{π}{3}) + 3 x : 6 (\frac{1}{2} cos (2 x) + \frac{3}{2} sin (2 x)) + 3 x

= \frac{d}{d x} (6 (\frac{1}{2} cos (2 x) + \frac{3}{2} sin (2 x)) + 3 x)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (6 (\frac{1}{2} cos (2 x) + \frac{3}{2} sin (2 x))) + \frac{d}{d x} (3 x)

\frac{d}{d x} (6 (\frac{1}{2} cos (2 x) + \frac{3}{2} sin (2 x))) = 6 (- sin (2 x) + 3 cos (2 x))

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

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

n = 5 \sum \infty \frac{1}{4 ^{n} + 1}

\frac{z}{( z + 1 ) ^{2}}

8 cos^{3} (t)

f (x) = \frac{4 x}{( x ^{2} + 1 )}

\int x^{- \frac{1}{4}} d x

What is the derivative of f(x)=6cos(2x-pi/3)+3x ?
The derivative of f(x)=6cos(2x-pi/3)+3x is 6(-sin(2x)+sqrt(3)cos(2x))+3
What is the first derivative of f(x)=6cos(2x-pi/3)+3x ?
The first derivative of f(x)=6cos(2x-pi/3)+3x is 6(-sin(2x)+sqrt(3)cos(2x))+3