# Thread: Product Rule

1. ## Product Rule

I'm new to all this stuff so bear with me...

I'm given $y =2x(8x+6)$

I need to find $y'$

So i do it...

$\frac {dx}{dy} = u\frac{dv}{dx} + v\frac {du}{dx}$

$2x.\frac{d(8x+6)}{dx} + (8x+6).\frac {d2x}{dx}$

$16x+12 + 16x +12$

$32x +24$

But when I check on my calculator it says $32x+12$

2. Originally Posted by jgv115
I'm new to all this stuff so bear with me...

I'm given $y =2x(8x+6)$

I need to find $y'$

So i do it...

$\frac {dx}{dy} = u\frac{dv}{dx} + v\frac {du}{dx}$

$2x.\frac{d(8x+6)}{dx} + (8x+6).\frac {d2x}{dx}$

$16x+12 + 16x +12$

$32x +24$

But when I check on my calculator it says $32x+12$

$y = 2x(8x+6)$.

$\frac{dy}{dx} = u\,\frac{dv}{dx} + v\,\frac{du}{dx}$

$= 2x(8) + (8x + 6)(2)$

$= 16x + 16x + 12$

$= 32x + 12$.

3. Hey that's exactly what my book says! But I always get different.

$= 2x(8) + (8x + 6)(2)$

I don't understand how you get $2x(8)$

How can $\frac{d(8x+6)}{dx}$ become 8?

4. Originally Posted by jgv115
Hey that's exactly what my book says! But I always get different.

$= 2x(8) + (8x + 6)(2)$

I don't understand how you get $2x(8)$

How can $\frac{d(8x+6)}{dx}$ become 8?
The derivative of $8x$ is $8$ and the derivative of $6$ is $0$.

So the derivative of $8x + 6$ is $8 + 0 = 8$.

5. oohh!!!!!!!! Thanks so much!!!!!!!

If I have $\frac {d(7x+6)}{dx}$

Then would it be $7$??

6. Originally Posted by jgv115
oohh!!!!!!!! Thanks so much!!!!!!!

If I have $\frac {d(7x+6)}{dx}$

Then would it be $7$??
Correct.

7. Thanks Prove It!! I appreciate the help!