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Thread: Differentiating this Expression

  1. #1
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    Differentiating this Expression

    Can anyone help me with this? The question asks to differentiate the following...

    (x-5)^8 * (x+3)^6

    The way I attempted it was to try using the product rule, but I don't get the same answer as stated in the book.

    I used the chain rule to differentiate each of the two arguments, getting 8(x-5)^7 and 6(x+3)^5 respectively. At that point I attempted to use the product rule, getting:

    8(x-5)^7*(x+3)^6 + (x-5)^8*6(x+3)^5

    but this is not the correct answer, unless there's some algebra I'm not seeing with the answer above - I suspect I'm doing something amiss. Any help is appreciated!
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  2. #2
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    Re: Differentiating this Expression

    Quote Originally Posted by YoungMarbleGiant View Post
    differentiate the following...
    $\displaystyle (x-5)^8 * (x+3)^6$
    I used the chain rule to differentiate each of the two arguments, getting 8(x-5)^7 and 6(x+3)^5 respectively. At that point I attempted to use the product rule, getting:
    $\displaystyle 8(x-5)^7*(x+3)^6 + (x-5)^8*6(x+3)^5$
    but this is not the correct answer
    That answer is certainly correct. Why do you doubt it?
    It can be factored: $\displaystyle (x-5)^7(x+3)^5[8(x+3)+6(x-5)]$
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  3. #3
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    Re: Differentiating this Expression

    Quote Originally Posted by Plato View Post
    That answer is certainly correct. Why do you doubt it?
    It can be factored: $\displaystyle (x-5)^7(x+3)^5[8(x+3)+6(x-5)]$
    I was sure the method was correct, but the answer in the exercise book is quite different. It gives the answer as $\displaystyle 2(7x-3)(x-5)^7(x+3)^5$.

    I looked at ways of simplifying/factoring, but the $\displaystyle (7x-3)$ was throwing me off. Do you have any idea how to get to this answer from the point where I reached? Thank you.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Differentiating this Expression

    That's exactly what Plato said, because $\displaystyle [8(x+3)+6(x-5)]=(8x+24+6x-30)=14x-6=2(7x-3)$
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  5. #5
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    Re: Differentiating this Expression

    Alright, thanks to you both.
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