# Math Help - differentiation problem

1. ## differentiation problem

Differentiate for x

(3x^2-4x+3)sin(2x) / 9

(6x-x^-1/2+3)2cos(2x)/9

am i on the right track -

2. Not yet, you need to employ the product rule here which is, if $\displaystyle y= vu \implies y'=vu'+uv'$

Do you follow?

3. ## Nope

Barely touched calculus- derived instantaneous gradient and looked at x^4 becomes 4x^3, and table of derivatives. But i'm a real newbie.

4. Originally Posted by stophe73
Differentiate for x

(3x^2-4x+3)sin(2x) / 9

This guy here is the product rule

$\displaystyle y= vu \implies y'=vu'+uv'$

used when you differentiate a product.

In your case make $\displaystyle v= \frac{(3x^2-4x+3)}{9}, u =\sin(2x)$ ifnd $u'$ and $v'$

so we get,

$
\displaystyle \left(\frac{(3x^2-4x+3)\sin(2x)}{9}\right)' = \frac{(3x^2-4x+3)}{9}(\sin(2x))'}+\left(\frac{(3x^2-4x+3)}{9}\right)'\sin(2x)$

5. Thats a big Help.....thanks.