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Math Help - differentiation of trig functions

  1. #1
    Newbie Tom G's Avatar
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    differentiation of trig functions

    Hi, i'm struggling on this question and would appreciate some hints as to what I do next.
    Q: Given that x=tan(y/2), prove that dy/dx=2/1+x^2

    so far i have got this;

    dx/dy=[sec^2(y/2)]/2 dy/dx= 2/[cos^2(y/2)] = 2/1-sin^2(y/2)

    i think i need to change sin^2(y/2) to tan^2(y/2) which will give me the x^2, unfortunately i can't think how. presumably the minus sign i have will become a plus as you can't square x and get a negative answer.
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  2. #2
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    Quote Originally Posted by Tom G View Post
    Hi, i'm struggling on this question and would appreciate some hints as to what I do next.
    Q: Given that x=tan(y/2), prove that dy/dx=2/1+x^2

    so far i have got this;

    dx/dy=[sec^2(y/2)]/2 dy/dx= 2/[cos^2(y/2)] = 2/1-sin^2(y/2)

    i think i need to change sin^2(y/2) to tan^2(y/2) which will give me the x^2, unfortunately i can't think how. presumably the minus sign i have will become a plus as you can't square x and get a negative answer.
    dx/dy=[sec^2(y/2)]/2 is good (there's NO dy/dx at the end btw). But forget about the 2/[cos^2(y/2)] = 2/1-sin^2(y/2) business .....

    dx/dy=[sec^2(y/2)]/2 therefore dy/dx = 2/[sec^2(y/2)].

    Now remember the standard identity tan^2(y/2) + 1 = sec^2(y/2).

    But x=tan(y/2).

    Therefore dy/dx = ....
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tom G View Post
    Hi, i'm struggling on this question and would appreciate some hints as to what I do next.
    Q: Given that x=tan(y/2), prove that dy/dx=2/1+x^2

    so far i have got this;

    dx/dy=[sec^2(y/2)]/2 dy/dx= 2/[cos^2(y/2)] = 2/1-sin^2(y/2)

    i think i need to change sin^2(y/2) to tan^2(y/2) which will give me the x^2, unfortunately i can't think how. presumably the minus sign i have will become a plus as you can't square x and get a negative answer.
    i'd approach the problem like this:

    x = \tan \frac y2

    \Rightarrow \arctan x = \frac y2

    \Rightarrow y = 2 \arctan x

    \Rightarrow \frac {dy}{dx} = \frac 2{1 + x^2}

    as desired
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