Derivative of tan(x) with Examples and Chain Rule

Here, we learn how to find the derivative of tan(x) step by step. Using the trigonometric identity \( \tan x = \dfrac{\sin x}{\cos x} \), together with the known derivatives of \( \sin x \) and \( \cos x \), we apply the quotient rule to derive the formula \( \dfrac{d}{dx}\tan x = \sec^2 x \). We also extend this result to composite functions of the form \( \tan(u(x)) \) using the chain rule, and work through several examples to reinforce understanding.

Calculation of the Derivative of tan x

A trigonometric identity relating \( \tan x \), \( \sin x \) and \( \cos x \) is given by \[ \tan x = \dfrac { \sin x }{ \cos x } \] One way to find the derivative of \( \tan x \) is to use the quotient rule of differentiation; hence \[ {\dfrac {d}{dx} \tan x = \dfrac {d}{dx} (\dfrac{\sin x }{\cos x}) = \dfrac{{ (\dfrac {d}{dx}\sin x) }{ \cos x } - \sin x (\dfrac {d}{dx} \cos x) }{\cos^2 x}} \] Use the formulae for the derivative of the trigonometric functions \( \sin x \) and \( \cos x \) given by \( \dfrac {d}{dx}\sin x = \cos x \) and \( \dfrac {d}{dx}\cos x = - \sin x \) and substitute to obtain \[ {\dfrac {d}{dx} \tan x = (\dfrac{{ \cos x \cos x } - \sin x (-\sin x) }{\cos^2 x}} \] Simplify \[ = \dfrac{ \cos^2 x + \sin^2 x } {\cos^2 x} = \dfrac{ 1 }{\cos^2 x} = \sec^2 x \]
conclusion
\[ \displaystyle {\dfrac {d}{dx} \tan x = \sec^2 x} \]

Graph of tan x and its Derivative

The graphs of \( \tan(x) \) and its derivative are shown below.

Graph of tan x and its derivative

Derivative of the Composite Function tan (u(x))

We now have a composite function which is a function (tan) of another function (u). Use the chain rule of differentiation to write

\( \displaystyle \dfrac{d}{dx} \tan (u(x)) = (\dfrac{d}{du} \tan u) (\dfrac{d}{dx} u ) \)

Simplify

\( = \sec^2 u \dfrac{d}{dx} u \)

Conclusion

\[ \displaystyle \dfrac{d}{dx} \tan (u(x)) = \sec^2 u \dfrac{d}{dx} u \]

Example 1
Find the derivative of the composite tan functions

  1. \( f(x) = \tan (2x -1) \)
  2. \( g(x) = \tan (\cos(x)) \)
  3. \( h(x) = \tan (\dfrac{x-1}{x+2}) \)

Solution to Example 1


  1. Let \( u(x) = 2x - 1 \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} (2x - 1) = 2 \) and apply the rule

    \( \displaystyle \dfrac{d}{dx} f(x) = (\sec^2 u \dfrac{d}{dx} u = \sec^2 (2x-1) \times 2 = 2 \sec^2 (2x-1) ) \)


  2. Let \( u(x) = \cos x \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} \cos x = - \sin x \) and apply the rule

    \( \displaystyle \dfrac{d}{dx} g(x) = (\sec^2 u \dfrac{d}{dx} u = \sec^2 (\cos x) \times (- \sin x) = - \sin x \sec^2 (\cos x) \)


  3. Let \( u(x) = \dfrac{x-1}{x+2} \) and therefore \( \dfrac{d}{dx} u = \dfrac{3}{(x+2)^2} \) and apply the rule obtained above

    \( \displaystyle \dfrac{d}{dx} h(x) = \sec^2 u \dfrac{d}{dx} u = \sec^2 (\dfrac{x-1}{x+2}) \times \dfrac{3}{(x+2)^2} = \dfrac {3 \sec^2 (\dfrac{x-1}{x+2})}{(x+2)^2} \)


More References and links

Rules of Differentiation of Functions in Calculus.
Trigonometric Identities and Formulas.
Derivatives of the Trigonometric Functions.
Chain Rule of Differentiation in Calculus.