Derivative of Tangent Function/Proof 2
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Theorem
- $\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x = \dfrac 1 {\cos^2 x}$
when $\cos x \ne 0$.
Proof
\(\ds \map {\frac \d {\d x} } {\tan x}\) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\map \tan {x + h} - \tan x} h\) | Definition of Derivative of Real Function at Point | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h} {1 - \tan x \tan h} - \tan x} h\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h - \tan x + \tan^2 x \tan h} {1 - \tan x \tan h} } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\tan h + \tan^2 x \tan h} {h \paren {1 - \tan x \tan h} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {1 + \tan^2 x} {1 - \tan x \tan h} \cdot \lim_{h \mathop \to 0} \frac {\tan h} h\) | Product Rule for Limits of Real Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 + \tan^2 x} {1 - \tan x \tan 0} \cdot 1\) | Limit of $\dfrac {\tan x} x$ at Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \tan^2 x\) | Tangent of Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \sec^2 x\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\cos^2 x}\) | Secant is Reciprocal of Cosine ($\cos x \ne 0$) |
$\blacksquare$