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 |
$\blacksquare$