Section 3.7 : Derivatives of Inverse Trig Functions
5. Differentiate \(\displaystyle h\left( x \right) = \frac{{{{\sin }^{ - 1}}\left( x \right)}}{{1 + x}}\) .
Show SolutionNot much to do here other than take the derivative using the formulas from class.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{h'\left( x \right) = \frac{{\frac{{1 + x}}{{\sqrt {1 - {x^2}} }} - {{\sin }^{ - 1}}\left( x \right)}}{{{{\left( {1 + x} \right)}^2}}} = \frac{{1 + x - \sqrt {1 - {x^2}} {{\sin }^{ - 1}}\left( x \right)}}{{\sqrt {1 - {x^2}} {{\left( {1 + x} \right)}^2}}}}}\]