(arctanx)' = ?
\tan{ (\arctan{(x)} ) } = x \quad | \quad \frac{d}{dx} \\\\ \left[ 1 + \tan^2{( \arctan{(x)} )} \right] \times [\arctan{(x)} ]' = 1 \right( \\\\ \left( 1 + x^2 \right) \times [\arctan{(x)} ]' = 1 \right)\\\\ \boxed{[\arctan{(x)} ]' = \frac{1}{ 1 + x^2 } \right)}
Here are the derivatives of the inverse trig functions.........http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx
(arctanx)' = ? (arcsinx)' = ?
(arcsin(x))′= ?sin(arcsin(x))=x|ddxcos(arcsin(x))∗(arcsin(x))′=1
\small{ (\arcsin(x))' = \frac{1} { cos{ ( \arcsin{(x)} ) } } \quad | \quad cos{ ( \arcsin{(x)} ) } = \sqrt{ 1- sin^2{ ( \arcsin{(x)} ) } } = \sqrt{ 1- x^2 } } }\\\\ \boxed{(\arcsin(x))' = \frac{1} { \sqrt{ 1- x^2 } }}
(arctanx)' = ?
\tan{ (\arctan{(x)} ) } = x \quad | \quad \frac{d}{dx} \\\\ \left[ 1 + \tan^2{( \arctan{(x)} )} \right] \times [\arctan{(x)} ]' = 1 \right( \\\\ \left( 1 + x^2 \right) \times [\arctan{(x)} ]' = 1 \right)\\\\ \boxed{[\arctan{(x)} ]' = \frac{1}{ 1 + x^2 } \right)}