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  1. #1
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    derivatives

    what is the first derivative of

    sec^2 x ...

    thanks =)
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  2. #2
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    Quote Originally Posted by go.leafs.go.89 View Post
    what is the first derivative of

    sec^2 x ...

    thanks =)
    2sec^2xtanx
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  3. #3
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    so the derivative of sec x is tan x?
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  4. #4
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    Quote Originally Posted by go.leafs.go.89 View Post
    what is the first derivative of



    sec^2 x ...

    thanks =)
    let g(x)=\sec^{2}(x)


    f(u)=u^2 \mbox{ and }u=\sec(x)

    f'(u)=2u \mbox{ and } u'=\sec(x)\tan(x)

    then g(x)=f(u)

    taking the derivative by the chain rule gives

    g'(x)=f'(u)\cdot u'=2u \cdot (\sec(x)\tan(x))

    but u is u=\sec(x)

    so we get

    g'(x)=f'(u)\cdot u'=2u \cdot (\sec(x)\tan(x))=2\sec(x) \cdot (\sec(x)\tan(x))=2\sec^{2}(x)\tan(x)
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  5. #5
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    hmm
    i had to use that in a question

    lim tan x - x
    x->0 ---------
    x^2

    i used l'hopital's rule to do this .. but couldn't get very far ..
    any suggestions?
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by go.leafs.go.89 View Post
    hmm

    i had to use that in a question

    lim tan x - x
    x->0 ---------
    x^2

    i used l'hopital's rule to do this .. but couldn't get very far ..
    any suggestions?


    \lim_{x \to 0}\frac{\tan(x)-x}{x^2}=

    by L.H.

    \lim_{x \to 0}\frac{\sec^{2}(x)-1}{2x}

    this is of the from 0/0 so by L.H again

    \lim_{x \to 0}\frac{2\sec^2(x)\tan(x)}{2}=\frac{2(1)\cdot 0}{2}=0

    I hope this helps
    Last edited by TheEmptySet; April 27th 2008 at 04:53 PM. Reason: left off a square
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    \lim_{x \to 0}\frac{\tan(x)-x}{x^2}=

    by L.H.

    \lim_{x \to 0}\frac{\sec(x)-1}{2x}

    this is of the from 0/0 so by L.H again

    \lim_{x \to 0}\frac{\sec^2(x)\tan(x)}{2}=\frac{1\cdot 0}{2}=0

    I hope this helps
    yes it does .. thanks a lot
    however .. when you first differentiated

    tanx-x
    ------
    x^2

    isn't the derivative of tanx = sec^2x and not sec x
    which would lead up to having the 2nd derivative = 2sec^2xtanx

    ?
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  8. #8
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    Quote Originally Posted by go.leafs.go.89 View Post
    yes it does .. thanks a lot
    however .. when you first differentiated

    tanx-x
    ------
    x^2

    isn't the derivative of tanx = sec^2x and not sec x
    which would lead up to having the 2nd derivative = 2sec^2xtanx

    ?
    Yes it is a typo I but the limit is still zero.

    I will fix my post above
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  9. #9
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    thanks so much!
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