Derivative of an inverse trig function

  • #1

Homework Statement



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

Homework Equations



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

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???
 

Answers and Replies

  • #2
430
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]
 
  • #3
ooooooooh ok. that makes more sense. thats clever hehehe

thank you so much :smile:
 
  • #4
Gib Z
Homework Helper
3,346
6
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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