Calculus Quiz 2

Some authors define $y=\sec^{-1}x \Leftrightarrow \sec y=x$ and $y \in \Bigl[\left[0, {{\pi} \over {2}}\right) \cup \left({{\pi} \over {2}}, \pi \right]\Bigr]$.

Show that with this definition we have

$${{d} \over {dx}}\left(\sec^{-1}x\right)= {{1} \over {|x|\sqrt{x^2-1}}}, |x| \gt 1$$

Sol)

Let $y=\sec^{-1}x$ where $|x| \gt 1$

Then,

$ y=\sec^{-1}x$ 

$\implies x=\sec y$ 

$\implies {{dx} \over {dy}}= \sec y \tan y \rightarrow \sec y \tan y={{\sin y} \over {\cos^2 y}} \rightarrow$ 위에 주어진 범위에서 $\sin y \gt 0$       $\therefore \sec y \tan y \gt 0$ 

$\implies {{dy} \over {dx}} = {{1} \over {\sec y \tan y}}$ 

$\implies \left({{dy} \over {dx}}\right)^2 = {{1} \over {\sec^2 y \tan^2 y}} = {{1} \over {\sec^2 y \left(\sec^2-1\right)}}={{1} \over {x^2 \left(x^2-1\right)}}$

$\implies \left|{{dy} \over {dx}}\right|={{1} \over {|x| \sqrt{x^2-1}}}$

$$\therefore {{d} \over {dx}}\left(\sec^{-1}x\right)= {{1} \over {|x|\sqrt{x^2-1}}}, |x| \gt 1$$