Derivative of Square of Tangent

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Theorem

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

when $\cos x \ne 0$.


Proof

\(\ds \map {\dfrac \d {\d x} } {\tan^2 x}\) \(=\) \(\ds 2 \tan x \map {\dfrac \d {\d x} } {\tan x}\) Chain Rule for Derivatives, Derivative of Square Function
\(\ds \) \(=\) \(\ds 2 \tan x \sec^2 x\) Derivative of Tangent Function

$\blacksquare$