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Math Help - Find derivative using product rule....

  1. #1
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    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)
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  2. #2
    Super Member Aryth's Avatar
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    Well, we have:

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

    We can say that:

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

    We know that:

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

    So we just plug in and simplify:

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

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

    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:

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

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

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

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

    They match exactly.
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  3. #3
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    Quote Originally Posted by eniav View Post
    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:

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

    Let f(b) = f(x)g(x), where:

    f(x) = x^2-4x+2

    and

    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.
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