Derivative of Tangent Function
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Theorem
- $\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x = \dfrac 1 {\cos^2 x}$
when $\cos x \ne 0$.
Corollary 1
- $\map {\dfrac \d {\d x} } {\tan a x} = a \sec^2 a x$
Corollary 2
- $\dfrac \d {\d x} \tan x = 1 + \tan^2 x$
Corollary 3
- $\map {\dfrac \d {\d x} } {\tan a x} = a \paren {\tan^2 a x + 1}$
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:
- $\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
Then:
\(\ds \map {\dfrac \d {\d x} } {\tan x}\) | \(=\) | \(\ds \frac {\cos x \cos x - \sin x \paren {-\sin x} } {\cos^2 x}\) | 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.
$\blacksquare$
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 $\dfrac {\tan x} x$ 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 ($\cos x \ne 0$) |
$\blacksquare$
Also see
Sources
- 1944: R.P. Gillespie: Integration (2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $6$.
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation
- 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $3.$ Trigonometric functions
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions: Trigonometric functions
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $6$: Derivatives