With all due respect to tan's inverse

Calculus Level 3

Find the derivative of \(\displaystyle \tan ^{ -1 }{ \left( \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) } \) with respect to \(\displaystyle \tan ^{ -1 }{ x }\), when \(x\neq 0\).

Notation: \(\displaystyle \tan ^{ -1 }{ y } =\arctan { y }\).

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