Derivative of an inverse trig function

  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:

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

    which works out as:

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

    and:

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

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

    [tex]\frac{-1}{\sqrt{1-t^{2}}}[/tex]

    what did I do wrong???
     
  2. jcsd
  3. 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. ooooooooh ok. that makes more sense. thats clever hehehe

    thank you so much :smile:
     
  5. Gib Z

    Gib Z 3,348
    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.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook