1. ## Product Rule

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

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

I need to find $\displaystyle y'$

So i do it...

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

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

$\displaystyle 16x+12 + 16x +12$

$\displaystyle 32x +24$

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

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

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

I need to find $\displaystyle y'$

So i do it...

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

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

$\displaystyle 16x+12 + 16x +12$

$\displaystyle 32x +24$

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

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

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

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

$\displaystyle = 16x + 16x + 12$

$\displaystyle = 32x + 12$.

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

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

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

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

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

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

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

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

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

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

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

Then would it be $\displaystyle 7$??

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

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

Then would it be $\displaystyle 7$??
Correct.

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