2 Answers | Add Yours
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)
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
We’ve answered 319,199 questions. We can answer yours, too.Ask a question