Derivatives: rules and functions

Derivative of tan x

The derivative of tan⁡x\tan x is 1cos⁡2x\frac{1}{\cos^2 x}.

(tan⁡x)′=1cos⁡2x(\tan x)'=\frac{1}{\cos^2 x}

Apply the quotient rule to tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}. The numerator becomes cos⁡2x+sin⁡2x=1\cos^2 x+\sin^2 x=1.

(tan⁡x)′=cos⁡x⋅cos⁡x−sin⁡x⋅(−sin⁡x)cos⁡2x=1cos⁡2x\begin{gathered} (\tan x)' \\[4pt] =\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^2 x} \\[4pt] =\frac{1}{\cos^2 x} \end{gathered}

You can memorize the result, but derive it by hand once so it sticks.

Examples

  1. 01Differentiateddxtan⁡x\displaystyle \frac{d}{dx} \tan xAnswerHide
    Answer1cos⁡2x\displaystyle \frac{1}{\cos^2 x}
  2. 02Differentiate:tan⁡x\displaystyle \tan xAnswerHide
    Answer1cos⁡2x\displaystyle \frac{1}{\cos^{2}x}
  3. 03What is the value?cos⁡2x+sin⁡2x\displaystyle \cos^{2}x+\sin^{2}xAnswerHide
    Answer1\displaystyle 1
  4. 04What is the value at x=0?(tan⁡x)′=1cos⁡2x\displaystyle (\tan x)'=\frac{1}{\cos^{2}x}AnswerHide
    Answer1\displaystyle 1
  5. 05Expand:cos⁡xcos⁡x−sin⁡x(−sin⁡x)\displaystyle \cos x\cos x-\sin x(-\sin x)AnswerHide
    Answercos⁡2x+sin⁡2x\displaystyle \cos^{2}x+\sin^{2}x
  6. 06What goes in □?tan⁡x=sin⁡x□\displaystyle \tan x = \frac{\sin x}{\square}AnswerHide
    Answercos⁡x\displaystyle \cos x

FAQ

Q1How do you derive (tan⁡x)′(\tan x)'?Show answerHide
A
Apply the quotient rule to tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}. The numerator becomes cos⁡2x+sin⁡2x=1\cos^2 x+\sin^2 x=1, giving 1cos⁡2x\frac{1}{\cos^2 x}.

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Published 2026-09-03 · Updated 2026-09-05