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Math Help - derivatives of trigonometric functions

  1. #1
    rmn
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    derivatives of trigonometric functions

    find dy/dx of

    y= (cosx/x) + (x/cosx)

    i got up to

    ((xsinx-cosx)/x^2 ) + ((cosx+sinx)/cosx^2))

    and i don't know how to simplify it.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rmn View Post
    find dy/dx of

    y= (cosx/x) + (x/cosx)

    i got up to

    ((xsinx-cosx)/x^2 ) + ((cosx+sinx)/cosx^2))

    and i don't know how to simplify it.
    perhaps it would be easier to simplify if you had combined the fractions first.

    note that: y = \frac {\cos^2 x + x^2}{x \cos x}

    Now, perform the quotient rule on that.
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  3. #3
    rmn
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    ok so I now have

    ((xcosx)(-sinx^2 + 2x) - (cosx^2+x^2)(x)(-sinx)+cosx) / (xcosx)^2

    i'm confused of how to simplify.. like how do you multiply xcosx with stuff?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rmn View Post
    ok so I now have

    ((xcosx)(-sinx^2 + 2x) - (cosx^2+x^2)(x)(-sinx)+cosx) / (xcosx)^2

    i'm confused of how to simplify.. like how do you multiply xcosx with stuff?
    that derivative is incorrect. you need the chain rule to derive things like \cos^2 x

    this is my 42th post!!!!
    Last edited by Jhevon; October 8th 2007 at 01:17 PM.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    after trying your way and my way, i found that the problem is generally easier to do your way. however, i do not see a worth-while way to simplify anything, so just leave the answer as:

    y' = \frac {-x \sin x - \cos x}{x^2} + \frac {\cos x + \sin x}{\cos^2 x}

    or if you feel like, you can combine these fractions to get: \frac {-x \sin x \cos^2 x - \cos^3 x + x^2 \cos x + x^2 \sin x}{x^2 \cos^2 x}

    but i don't see any motivation for doing even that
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