# mathDifferentiate y = 2sin(3x+1)cos(3x+1)

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We have to differentiate y = 2sin(3x+1)cos(3x+1)

y = 2sin(3x+1)cos(3x+1)

=> y = sin [2(3x + 1)]

=> y = sin (6x + 2)

Use the chain rule.

y' = cos (6x + 2) * 6

**The required derivative is 6*cos (6x + 2)**

This problem requires the product rule and the chain rule:

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

(u*v)' = u'*v + u*v'

Let u(x) = 2sin(3x+1) => u'(x) = 2[cos (3x+1)]*(3x+1)'

u'(x) = 6*cos (3x+1)

v(x) = cos(3x+1) => v'(x) = [-sin(3x+1)]*(3x+1)'

v'(x) = -3*sin(3x+1)

We'll substitute in the expression of product:

f'(x) = 6*cos (3x+1)*cos (3x+1) - 2sin(3x+1)*3*sin(3x+1)

f'(x) = 6[cos (3x+1)]^2 - 6[sin (3x+1)]^2

We'll factorize by 6:

f'(x) = 6[cos (3x+1)^2 - sin (3x+1)^2]

**f'(x) = 6cos 2(3x+1)**

** **

We can use here

**2 sin A cos B = sin(A+B) + sin(A-B)**