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Derivative of inverse trig functions

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data

    ln(sec^-1(3x^2 +1))

    2. Relevant equations



    3. The attempt at a solution

    1/sec-1(3x2+1) * 1/(3x2+1)(sqrt(3x2+1)2-1) * 6x

    Is this correct ?, do I just simplify from here ?
     
  2. jcsd
  3. Apr 12, 2014 #2

    Fredrik

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    It looks wrong. There should be more trigonometric stuff in the result. How did you get that result? Do you know how to take the derivative of an inverse function: $$(f^{-1})'(x)=?$$
     
  4. Apr 12, 2014 #3
    If you mean:
    $$\frac{1}{\sec^{-1}(3x^2+1)}\frac{1}{(3x^2+1)\sqrt{(3x^2+1)^2-1}}(6x)$$
    then yes, it is correct. :smile:
     
  5. Apr 12, 2014 #4

    Dick

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    It's a little hard to interpret exactly without more parantheses, but it looks ok to me. What kind of 'more trignonometric' stuff are you looking for?
     
  6. Apr 12, 2014 #5

    Fredrik

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    I didn't try to work it all out, but I'm thinking that
    $$\frac{d}{dx}\sec^{-1}(f(x))=(\sec^{-1})'(f(x))f'(x)=\frac{1}{\sec'(\sec^{-1}(f(x)))}f'(x)$$ and
    $$\sec'(x)=\frac{d}{dx}\frac{1}{\cos x}=-\frac{1}{\cos^2x}(-\sin x).$$ So it looks like we get a big mess of "trigonometric stuff". But since you're both saying that he's right, I assume that I'm missing something.
     
  7. Apr 12, 2014 #6

    Dick

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    I'd do it the other way around. Since sec(arcsec(x))=x, sec'(arcsec(x))*arcsec'(x)=1. sec'=sec*tan and tan(arcsec(x))=sqrt(x^2-1). No trig left in the end.
     
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