Derivative of an inverse trig function

[efekwulsemmay]

Homework Statement

y=sec^{-1}\frac{1}{t}, 0<t<1

Homework Equations

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

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:

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

which,when I substituted for u, I got:

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

which works out as:

=\frac{-1}{\left|\frac{1}{t}\right|\cdot t^{2}\cdot\sqrt{\frac{1^{2}}{t^{2}}-1}}

and:

=\frac{-1}{t\cdot\sqrt{1-t^{2}}}

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

\frac{-1}{\sqrt{1-t^{2}}}

what did I do wrong?

 

[rasmhop]

The step,
\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}}}
is incorrect. Remember
a\sqrt{b} = \sqrt{a^2b}
so you have,
t\sqrt{1/t^2-1} = \sqrt{1-t^2}

 

[efekwulsemmay]

ooooooooh ok. that makes more sense. that's clever hehehe

thank you so much

 

[Gib Z]

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