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Math Help - Constant Multiple Rule...Yes or No?

  1. #1
    Senior Member nycmath's Avatar
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    Constant Multiple Rule...Yes or No?

    Find s'(t).

    Does the constant multiple rule apply here?

    s(t) = [-2(2-t) * sqrt{1+t}]/3

    I pulled out 1/3 but my answer is nothing like the book's answer.

    Applying the constant rule, the problem now looks like this:

    s(t) = (1/3) times [-2(2-t) * sqrt{1+t}]

    The fraction (1/3) comes along for the ride.
    I then know to apply the product rule to everything else.

    The answer is (t)/sqrt{1+t}. Can someone work out this problem for me? I would love to see the steps. I never post a question in this forum without working out the problem many times on my own. Please, help. Thank you.
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  2. #2
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    Re: Constant Multiple Rule...Yes or No?

    Yes the constant multiple rule applies here, in fact, it would be easiest if you take out -2/3 as a factor.
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    Senior Member nycmath's Avatar
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    Re: Constant Multiple Rule...Yes or No?

    I will try this on my own one more time. I never like giving up.
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    Re: Constant Multiple Rule...Yes or No?

    My answer:

    (-t)/[2(sqrt{1+t})

    The book's answer:

    (t)/sqrt{1+t}

    I came close but not right. Can you solve this one step by step? Thanks again.
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    Re: Constant Multiple Rule...Yes or No?

    Ok, so you apply the product rule to -2(2-t)\sqrt{1+t}.

    You should get \sqrt{1+t}\frac{d}{dt}(-2(2-t))-2(2-t)\frac{d}{dt}\sqrt{1+t}. The first term you can multiply top and bottom by \sqrt{1+t} to get \sqrt{1+t} in the denominator, and the second will have it when you take the derivative. So what does each term have in the numerator?

    - Hollywood
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    Senior Member nycmath's Avatar
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    Re: Constant Multiple Rule...Yes or No?

    I'll try this one again tomorrow (my day off). I am going chapter by chapter in my single variable calculus text which covers calculus 1 and 2. I always wanted to learn calculus 1-3. Why take out a loan and run myself into debt?

    The internet makes it possible for me to learn calculus.
    If I can go as far as mutivariable calculus on my own and learn it sufficiently well, it is a great accomplishment. The next chapter is Implicit Differentiation.
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