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.