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Homework Help: Derivative of an inverse trig function

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]y=sec^{-1}\frac{1}{t}, 0<t<1[/tex]

    2. Relevant equations

    [tex]\frac{d}{dx}sec^{-1} x= \frac{1}{\left|x\right|\cdot\sqrt{x^{2}-1}}[/tex]

    3. The attempt at a solution
    Basically to simplify things I used u substitution so I let u=1/t then du/dt=-1/t2 and I got:

    [tex]y=sec^{-1}u\rightarrow y^{,}=\frac{1}{\left|u\right|\cdot\sqrt{u^{2}-1}}\cdot\frac{du}{dt}[/tex]

    which,when I substituted for u, I got:


    which works out as:

    [tex]=\frac{-1}{\left|\frac{1}{t}\right|\cdot t^{2}\cdot\sqrt{\frac{1^{2}}{t^{2}}-1}}[/tex]



    however, the answer as per the back of the book is:


    what did I do wrong???
  2. jcsd
  3. Feb 12, 2010 #2
    The step,
    [tex]\frac{-1}{\left|\frac{1}{t}\right|\cdot t^{2}\cdot\sqrt{\frac{1^{2}}{t^{2}}-1}} = \frac{-1}{t\cdot\sqrt{1-t^{2}}}[/tex]
    is incorrect. Remember
    [tex]a\sqrt{b} = \sqrt{a^2b}[/tex]
    so you have,
    [tex]t\sqrt{1/t^2-1} = \sqrt{1-t^2}[/tex]
  4. Feb 12, 2010 #3
    ooooooooh ok. that makes more sense. thats clever hehehe

    thank you so much :smile:
  5. Feb 12, 2010 #4

    Gib Z

    User Avatar
    Homework Helper

    For next time, you may remember that since [tex]\cos x = \frac{1}{\sec x}[/tex], it is true that [tex]\sec^{-1} \left(\frac{1}{x}\right) = \cos^{-1} x[/tex] and so the derivative is standard.
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