PART 13:反正割函數的微分

定理12

\(f(x) = {\rm{Se}}{{\rm{c}}^{ - 1}}x,\) ,則 \(f'(x) = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}) , \left| x \right| > 1\)

證明

\({\rm{(Se}}{{\rm{c}}^{ - 1}}x)' = {\left[ {{\rm{Co}}{{\rm{s}}^{ - 1}}\left( {1/x} \right)} \right]^\prime }\) ,使用連鎖律

\({\rm{(Se}}{{\rm{c}}^{ - 1}}x)' = \frac{{ - 1}}{{\sqrt {1 - {{\left( {1/x} \right)}^2}} }}{\left( {\frac{1}{x}} \right)^\prime }\)  

     \( = \frac{1}{{{x^2}\sqrt {1 - {{\left( {1/x} \right)}^2}} }}\)  

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

     \( = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}\) , \(\left| x \right| > 1\)