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Differentiation of Trig Functions

  1. Feb 22, 2006 #1

    Hootenanny

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    The question is as follows:
    Find [tex] \frac{dy}{dx} \; \; \; xtanx \; \; \; dx [/tex]
    The standard differential is given in the formula book as
    [tex] f(x) = \tan kx \Rightarrow f'(x) = k\sec^2 kx [/tex]
    Therefore, I got:
    [tex] \frac{dy}{dx} = x \sec^2 x [/tex]
    However, the answer given is
    [tex] \tan x + x \sec^2 x [/tex]
    I can't see where I've gone wrong, it seems like such a simple differential. Any help would be much appreciated.
     
    Last edited: Feb 22, 2006
  2. jcsd
  3. Feb 22, 2006 #2
    Have you heard of differentiation of products? Namely,

    [tex]\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)[/tex]
     
  4. Feb 22, 2006 #3
    The [itex] x [/itex] isn't a constant that can be left out when you differentiate. You need to use the product rule on the two terms [itex] x [/itex] and [itex] \tan(x) [/itex]
     
  5. Feb 22, 2006 #4
    what the book gave has k as a constant, in your problem x is not a constant. you need to use the product rule.
     
  6. Feb 22, 2006 #5

    Hootenanny

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    Oh yes. I release what I've done now. Thank's.
     
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