Derivatives: rules and functions

Derivative of tan x

The derivative of tan⁡x\tan x is 1cos⁡2x\dfrac{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=\dfrac{\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. 01Differentiate:tan⁡x\displaystyle \tan xAnswerHide
    Answer1cos⁡2x\displaystyle \frac{1}{\cos^{2}x}

    Steps

    1. Quotient rule on tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}: the numerator collapses to cos⁡2x+sin⁡2x\cos^{2}x+\sin^{2}x
    2. cos⁡2x+sin⁡2xcos⁡2x\displaystyle \frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=1cos⁡2x\displaystyle = \frac{1}{\cos^{2}x}
    3. The denominator is the original cos⁡2x\cos^{2}x; not sin⁡2x\sin^{2}x
  2. 02What is the value?cos⁡2x+sin⁡2x\displaystyle \cos^{2}x+\sin^{2}xAnswerHide
    Answer1\displaystyle 1

    Steps

    1. This holds for every x
    2. sin⁡2x+cos⁡2x\displaystyle \sin^{2}x+\cos^{2}x=1\displaystyle = 1
    3. At x=0x=0 for instance, sin⁡0=0\sin 0=0 and cos⁡0=1\cos 0=1, and 0+1=10+1=1
  3. 03What is the value at x=0?(tan⁡x)′=1cos⁡2x\displaystyle (\tan x)'=\frac{1}{\cos^{2}x}AnswerHide
    Answer1\displaystyle 1

    Steps

    1. Put in cos⁡0=1\cos 0 = 1
    2. 112\displaystyle \frac{1}{1^{2}}=1\displaystyle = 1
    3. Slope 1; the graph of tan⁡x\tan x leaves the origin at 45 degrees
  4. 04What goes in □?tan⁡x=sin⁡x□\displaystyle \tan x = \frac{\sin x}{\square}AnswerHide
    Answercos⁡x\displaystyle \cos x

    Steps

    1. This is what tan⁡\tan means
    2. tan⁡x\displaystyle \tan x=sin⁡xcos⁡x\displaystyle = \frac{\sin x}{\cos x}
    3. At x=π4x=\dfrac{\pi}{4}, for instance, sin⁡\sin and cos⁡\cos are equal so the quotient is 1; matching tan⁡π4=1\tan\dfrac{\pi}{4}=1
  5. 05Differentiateddxtan⁡x\displaystyle \frac{d}{dx} \tan xAnswerHide
    Answer1cos⁡2x\displaystyle \frac{1}{\cos^2 x}
  6. 06Expand: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

FAQ

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

Published 2026-09-03 · Updated 2026-10-14