1. ## Help: Product Differentiation!

Hi, Could someone give me somw much needed guidance on the following question:
Differentiate: f(x) = 2cos(3x - 1)/3x ?
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?

2. Originally Posted by joemc22
Hi, Could someone give me somw much needed guidance on the following question:
Differentiate: f(x) = 2cos(3x - 1)/3x ?
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?

Use the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}$

3. ## Product differentiation

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

4. Originally Posted by joemc22
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, (3x)^2 = 3^2 * x^2 = 9x^2. You must also square the three.
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
The function inside the brackets does not change when using the chain rule:

$f(x) = \frac{2cos(3x-1)}{3x}$

$u = 2cos(3x-1)$
$\frac{du}{dx} = -6sin(3x-1)$

$v = 3x$
$\frac{dv}{dx} = 3$

$f'(x) = \frac{vu'-uv'}{v^2} = \frac{-18xsin(3x-1)-6cos(3x-1)}{9x^2}$

5. Thanks,

Wasn't sure what to change the 2sin(3-1x) into, the self help lessons i'm learning from sometimes don't have the examples i need, but you help has made it clearer.

6. Originally Posted by danny arrigo
Use the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v + u v'}{v^2}$
Don't you mean $\left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}$

7. Originally Posted by Spec
Don't you mean $\left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}$
You're so right - how'd that negative turn to a postive. Must be the MHF gremlins