Derivative of Tangent Function/Proof 2

Theorem

$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x = \dfrac 1 {\cos^2 x}$

when $\cos x \ne 0$.

$\ds \map {\frac \d {\d x} } {\tan x}$

$=$

$\ds \lim_{h \mathop \to 0} \frac {\map \tan {x + h} - \tan x} h$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h} {1 - \tan x \tan h} - \tan x} h$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h - \tan x + \tan^2 x \tan h} {1 - \tan x \tan h} } h$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {\tan h + \tan^2 x \tan h} {h \paren {1 - \tan x \tan h} }$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {1 + \tan^2 x} {1 - \tan x \tan h} \cdot \lim_{h \mathop \to 0} \frac {\tan h} h$

$\ds $

$=$

$\ds \frac {1 + \tan^2 x} {1 - \tan x \tan 0} \cdot 1$

$\ds $

$=$

$\ds 1 + \tan^2 x$

$\ds $

$=$

$\ds \sec^2 x$

$\ds $

$=$

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

$\blacksquare$