Derivative of Tangent Function

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

when $\cos x \ne 0$.


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

Proof 1

From the definition of the tangent function:

$\tan x = \dfrac {\sin x} {\cos x}$

From Derivative of Sine Function:

$\map {\dfrac \d {\d x} } {\sin x} = \cos x$

From Derivative of Cosine Function:

$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$


\(\ds \map {\dfrac \d {\d x} } {\tan x}\) \(=\) \(\ds \frac {\cos x \cos x - \sin x \paren {-\sin x} } {\cos^2 x}\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {\cos^2 x + \sin^2 x} {\cos^2 x}\)
\(\ds \) \(=\) \(\ds \frac 1 {\cos^2 x}\) Sum of Squares of Sine and Cosine

This is valid only when $\cos x \ne 0$.

The result follows from the Secant is Reciprocal of Cosine.


Proof 2

\(\ds \map {\frac \d {\d x} } {\tan x}\) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map \tan {x + h} - \tan x} h\) 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 Tan X over X
\(\ds \) \(=\) \(\ds 1 + \tan^2 x\) 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 ($\cos x \ne 0$)


Also see