Derivative of Arctangent Function/Proof 1

Theorem

$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {1 + x^2}$

$\ds y$

$=$

$\ds \arctan x$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \tan y$

Definition of Real Arctangent

$\ds \leadsto \ \ $

$\ds \frac {\d x} {\d y}$

$=$

$\ds \sec^2 y$

$\ds $

$=$

$\ds 1 + \tan^2 y$

$\ds $

$=$

$\ds 1 + x^2$

Definition of $x$

$\ds \leadsto \ \ $

$\ds \frac {\d y} {\d x}$

$=$

$\ds \frac 1 {1 + x^2}$

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Inverse Ratios