Results 1 to 3 of 3

Math Help - differentia secx????????

  1. #1
    Member i_zz_y_ill's Avatar
    Joined
    Mar 2008
    Posts
    140

    differentia secx????????

    why is the differential of secx tanx when secx=/cosx=tanx/sinx

    im also confused as to why when you've got d(sec^(2)x)/dx)

    you can't do sec^(2)x=cos^(-2)x
    therefore dy/dx = 2cosxsinx=sin2x

    i think its wrong to say 1/cos^(2)x=cos^(-2)x but im not sure, is this the same as 1/arccosx im so confused,,,,,,,,,,,,,,,,,,,,,

    genreally just want to know how to diff secx >>>???? thnx,,,,but expaination as to y im wrong would be much appreciated thnx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by i_zz_y_ill View Post
    why is the differential of secx tanx when secx=/cosx=tanx/sinx

    im also confused as to why when you've got d(sec^(2)x)/dx)

    you can't do sec^(2)x=cos^(-2)x
    therefore dy/dx = 2cosxsinx=sin2x

    i think its wrong to say 1/cos^(2)x=cos^(-2)x but im not sure, is this the same as 1/arccosx im so confused,,,,,,,,,,,,,,,,,,,,,

    genreally just want to know how to diff secx >>>???? thnx,,,,but expaination as to y im wrong would be much appreciated thnx
    Your post is difficult to read and follow. It would benefit from better setting out.

    To differentiate y = sec x = 1/cos x, use the quotient rule.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi
    Quote Originally Posted by i_zz_y_ill View Post
    why is the differential of secx tanx when secx=/cosx=tanx/sinx
    I'm not sure to have understood what you mean. Anyway :

    \frac{\mathrm{d}}{\mathrm{d}x}(\sec x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{\c  os x}\right)<br />
=-\frac{-\sin x}{\cos^2x}=\sec x \tan x<br />

    im also confused as to why when you've got d(sec^(2)x)/dx)

    you can't do sec^(2)x=cos^(-2)x
    therefore dy/dx = 2cosxsinx=sin2x
    Because \frac{\mathrm{d}}{\mathrm{d}x}\left( u^n (x)\right)=n\cdot u'(x)\cdot u^{n-1}(x) which is not what you wrote. (-2-1=-3 )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Two First Order Differentia Questions
    Posted in the Differential Equations Forum
    Replies: 9
    Last Post: September 6th 2010, 03:33 AM
  2. Simplifying sin(x+4)secx?
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 2nd 2009, 06:38 AM
  3. integrate (secx)^4
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 15th 2008, 06:11 AM
  4. The reciprocal of secx
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 12th 2008, 07:30 PM
  5. How do I find Dx tan(x)^secx?
    Posted in the Calculus Forum
    Replies: 9
    Last Post: September 25th 2007, 11:27 PM

Search Tags


/mathhelpforum @mathhelpforum