# Thread: Find derivative using product rule....

1. ## Find derivative using product rule....

Thank You bobak (this is last one) tryed to multiply but didnt get right answer.

Find derivative using product rule

f(b) = (b^2 - 4b + 2)(2[b]^3 - b^2 + 4)

2. Well, we have:

$\displaystyle f(b) = (b^2 - 4b + 2)(2b^3 - b^2 + 4)$

We can say that:

$\displaystyle g(b) = b^2 - 4b + 2$
$\displaystyle h(b) = 2b^3 - b^2 + 4$

We know that:

$\displaystyle \frac{dy}{db}[g(b)h(b)] = g'(b)h(b) + h'(b)g(b)$

So we just plug in and simplify:

$\displaystyle f'(b) = (2b - 4)(2b^3 - b^2 + 4) + (6b^2 - 2b)(b^2 - 4b + 2)$

$\displaystyle f'(b) = (4b^4 - 2b^3 + 8b - 8b^3 + 4b^2 - 16) + (6b^4 - 24b^3 + 12b^2 - 2b^3 + 8b^2 - 4b)$

$\displaystyle f'(b) = 10b^4 -36b^3 + 24b^2 + 4b - 16$

The best way to verify this would be to multiply out the original equation above and use the power rule:

$\displaystyle f(b) = (b^2 - 4b + 2)(2b^3 - b^2 + 4)$

$\displaystyle f(b) = 2b^5 - b^4 + 4b^2 - 8b^4 + 4b^3 - 16b + 4b^3 - 2b^2 + 8$

$\displaystyle f(b) = 2b^5 -9b^4 + 8b^3 + 2b^2 - 16b + 8$

$\displaystyle f'(b) = 10b^4 - 36b^3 + 24b^2 + 4b - 16$

They match exactly.

3. Originally Posted by eniav
Thank You bobak (this is last one) tryed to multiply but didnt get right answer.

Find derivative using product rule

f(b) = (b^2 - 4b + 2)(2[b]^3 - b^2 + 4)
First, recall the product rule:

$\displaystyle \frac {d}{dx}\ [f(x)g(x)] = f(x)g'(x) + f'(x)g(x)$

Let $\displaystyle f(b) = f(x)g(x)$, where:

$\displaystyle f(x) = x^2-4x+2$

and

$\displaystyle g(x)=2x^3-x^2+4$

From this, you should be able to get your answer. To check your work, simply multiply out the original equation and find the derivative using the power rule.