# Thread: Product rule. (checking)

1. ## Product rule. (checking)

one more guys please

differentiate using the prduct rule

$
y(x) = (5x^3 + 8x^2 - 2x + 1)e^x
$

$
u = (5x^3 + 8x^2 - 2x + 1) v = e^x
$

$
\frac{du}{dx}= (10x^2 + 16x - 2)
$

so

$
\frac{dy}{dx}= v\frac{du}{dx} + u\frac{dv}{dx} = e^x(10x^2 + 16x - 2) + (5x^3 + 8x^2 - 2x + 1)e^x
$

2. Originally Posted by mxmadman_44
one more guys please

differentiate using the prduct rule

$
y(x) = (5x^3 + 8x^2 - 2x + 1)e^x
$
What are you doing?

$y'=(5x^3+8x^2-2x+1)'e^x+(5x^3+8x^2-2x+1)(e^x)'$

3. ## .

could you please show you workings how you gotto that answer i dont understand how you got v and u. I take it there is no need to differentiate anything. ie
$
3x^2 = 6x
$

4. The product rule states that:
$(uv)' = u'v+uv'$

Here we let $u=5x^3+8x^2-2x+1 \mbox{ and }v=e^x$

5. ## ..

okay got it, thanks!. I had a different equation but your answer looks right. Plus your know way more than me.

Cheers perfecthacker