# Find dy/dx for y = ( 4x - 1)(sin3x)Step by step process

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### 2 Answers

We have to find the derivative of y = ( 4x - 1)(sin3x)

y = ( 4x - 1)(sin3x)

=> y = 4x* sin 3x - sin 3x

Now use the product rules and the chain rule

dy/dx = [4x* sin 3x]' - [sin 3x]'

=> [4x]'* sin 3x + 4x*(sin 3x)' - [sin 3x]'

=> 4*sin 3x + 4x*3*cos 3x - 3*cos 3x

=> 4*sin 3x + 3*cos 3x*(4x - 1)

**The required derivative is 4*sin 3x + 3*cos 3x*(4x - 1)**

y= (4x-1)(sin3x)

We need to find dy/dx

We will use the product rule to find the derivative.

Let y= u*v such that:

u= 4x-1 ==> u' = 4

v= sin3x ==> v' = 3cos3x

Now we know that:

dy/dx = u'v + uv'

==> dy/dx = 4sin3x + 3cos3x(4x-1)

**==> dy/dx = 4sin3x + 12xcos3x - 3cos3x**