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)
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.
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.