Derivative of Tangent Function/Proof 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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

when $\cos x \ne 0$.


Proof

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

Retrieved from "https://proofwiki.org/w/index.php?title=Derivative_of_Tangent_Function/Proof_1&oldid=602215"