Hi, Could someone give me somw much needed guidance on the following question:
Differentiate: f(x) = 2cos(3x - 1)/3x ?
I'm struggling to start with this one.
Firstly does 2cos(3x - 1)/3x become: 2/3 xcos(1 - 3x), I'm trying to remove the division by 3x?
Can someone please give me some help so I can get on with this?
Thanks in advance.
OK, it's taking me a bit of time to get back into it!
So I have:
f(x)= 2cos(3x-1)/3x.
So u= 2cos(3x-1) and v= 3x
du/dx= 6sin(1-3x) and dv/dx= 3?
Using quotent rule I get:
(3x)6sin(1-3x) - 2cos(3x-1) x 3/3x^2, changing cos to sin and multiplying the last part by 3 I would then get the answer as 3x/3x^2
Does this look right, if not could you give me some pointers where i've gone wrong?
Thanks