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Math Help - simple derivatives?

  1. #1
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    Question simple derivatives?

    i absolutely LOVE derivatives. i CANNOT figure this one out. i have major trouble with square roots. >,<!!

    f(x)= 1/√(x+2),, so find f'(a)
    yeah. i know. simple right? i just can't figure out. i know its ^1/2 for the sqrt, but blehhhhh! i've tried this one a million times!

    i know the answer is -1/2(a+2)^3/2, back of the book. but erggg >.<

    well, while we're at it-->this one too?! just a simple derivation. it's just, "whaaa? why is THAT not right?" lol

    Last edited by mr fantastic; September 27th 2009 at 03:21 AM. Reason: Edited post title
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by jayjay60 View Post
    i absolutely LOVE derivatives. i CANNOT figure this one out. i have major trouble with square roots. >,<!!

    f(x)= 1/√(x+2),, so find f'(a)
    yeah. i know. simple right? i just can't figure out. i know its ^1/2 for the sqrt, but blehhhhh! i've tried this one a million times!

    i know the answer is -1/2(a+2)^3/2, back of the book. but erggg >.<

    well, while we're at it-->this one too?! just a simple derivation. it's just, "whaaa? why is THAT not right?" lol

    f(x)=\frac{1}{\sqrt{x+2}}=(x+2)^{-1/2}

    f'(x)=-\frac{1}{2}(x+2)^{-3/2}=-\frac{1}{2(x+2)^{3/2}}

    For the next one we need either the chain rule or a little algebra

    Here is the algebra first.

    F(x)=\left( \frac{1}{2}x\right)^4=\frac{1}{2^4}x^4=\frac{1}{16  }x^4


    F'(x)=\frac{1}{4}x^3

    Now by the chain rule we get

    F(x)=\left( \frac{1}{2}x\right)^4

    F(x)=4\left( \frac{1}{2}x\right)^3\cdot \left( \frac{1}{2}\right)=\frac{1}{4}x^3
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  3. #3
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    Talking

    thank you so much! i was getting 2x^3...i was putting 2x instead of 1/4...silly algebraic mistakes. they get me every time!
    i'm very grateful indeed!
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