# What is the derivative of the function y=2cos(3x-1)*sin(3x-1)?

### 2 Answers | Add Yours

The given function f(x) = 2*cos (3x-1)*sin(3x-1)

Use the product rule and chain rule to find the derivative.

f'(x) = 2*[cos (3x-1)]'*sin(3x-1) + 2*cos (3x-1)*[sin(3x-1)]'

=> 2*3*(-sin (3x - 1)*sin (3x - 1) + 2*cos (3x - 1)*3*sin (3x - 1)

=> 6[ (cos (3x - 1))^2 - (sin (3x - 1))^2]

=> 6*cos 2*(3x - 1)

=> 6*cos (6x - 2)

**The derivative of the given function is 6*cos (6x - 2)**

We recognize the formula of the double angle in the given expresison of the function:

y = 2 sin (3x-1)*cos (3x - 1)

y = sin 2*(3x-1)

y = sin (6x - 2)

Since the function to be differentiated is the result of composition of two functions, we'll use the chain rule to detemrine the first derivative of the function.

We'll differentiate with respect to x:

dy/dx = [cos (6x - 2)]*(6x - 2)'

dy/dx = 6*cos(6x - 2)

**The first derivative of the given function is: dy/dx = 6*cos(6x - 2).**