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Thread: Differentiating sec

  1. October 27th 2007, 06:03 AM #1
    curlywurlysqurly
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    Differentiating sec

    Hi, would someone please tell me how to differentiate sec^2 x?
    I tried it and got (-4sin2x)/(cos2x+1)^2
    Not sure if that is right tho
    xxx
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  2. October 27th 2007, 07:08 AM #2
    kalagota
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    Quote Originally Posted by curlywurlysqurly View Post
    Hi, would someone please tell me how to differentiate sec^2 x?
    I tried it and got (-4sin2x)/(cos2x+1)^2
    Not sure if that is right tho
    xxx
    differentiate?
    Dx[sec^2 x] = 2(sec x)(sec x tan x) = 2(sec^2 x)(tan x)
    you just nid to apply chain rule.
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  3. October 28th 2007, 03:14 AM #3
    curlywurlysqurly
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    sorry but i don't understand how you did that!!!
    where did the tan x come from???
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  4. October 28th 2007, 09:41 AM #4
    Krizalid
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    Do you know what chain rule is?

    f(x)=\sec^2x.

    To take its derivative, we use (u^n)'=nu^{n-1}\cdot u'.

    So, f'(x)=2\sec x\cdot(\sec x)'.

    Does that make sense?
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  5. October 28th 2007, 10:19 AM #5
    curlywurlysqurly
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    yea thats all fine, but how do u differentiate sec x? is it just a standard result or is there some way to do it?
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  6. October 28th 2007, 10:25 AM #6
    Krizalid
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    You know that \sec x=\frac1{\cos x}

    Now \Big[(\cos x)^{-1}\Big]'=-(\cos x)^{-2}\cdot(\cos x)'.

    So, \frac{\sin x}{\cos^2x}=\frac1{\cos x}\cdot\frac{\sin x}{\cos x}.
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  7. October 28th 2007, 11:03 AM #7
    curlywurlysqurly
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    okay thanks i get it now!
    thankyou all for your help!!!!
    xxx
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