Derivative of tan(x) with Examples and Chain Rule

Step-by-Step Proof, Graph, Chain Rule, and Solved Problems

Here, we learn how to find the derivative of tan(x) step by step. Using the trigonometric identity \( \tan x = \frac{\sin x}{\cos x} \), together with the known derivatives of \( \sin x \) and \( \cos x \), we apply the quotient rule to derive the formula \( \frac{d}{dx}\tan x = \sec^2 x \). 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:

\[ \tan x = \frac{\sin x}{\cos x} \]

One way to find the derivative of \( \tan x \) is to use the quotient rule of differentiation; hence:

\[ \frac{d}{dx} \tan x = \frac{d}{dx} \left(\frac{\sin x}{\cos x}\right) = \frac{\left(\frac{d}{dx}\sin x\right)\cos x - \sin x \left(\frac{d}{dx}\cos x\right)}{\cos^2 x} \]

Using the formulae for the derivatives of the trigonometric functions \( \sin x \) and \( \cos x \), given by \( \frac{d}{dx}\sin x = \cos x \) and \( \frac{d}{dx}\cos x = -\sin x \), we substitute to obtain:

\[ \frac{d}{dx} \tan x = \frac{\cos x \cos x - \sin x (-\sin x)}{\cos^2 x} \]

Simplify:

\[ = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \]
Conclusion: \[ \frac{d}{dx} \tan x = \sec^2 x \]

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 (tangent) of another function (\( u \)). Use the chain rule of differentiation to write:

\[ \frac{d}{dx} \tan(u(x)) = \left(\frac{d}{du} \tan u\right) \left(\frac{d}{dx} u\right) \]

Simplify:

\[ = \sec^2 u \cdot \frac{du}{dx} \]
Conclusion: \[ \frac{d}{dx} \tan(u(x)) = \sec^2 u \frac{du}{dx} \]

Examples with Solutions

Example 1: Find the derivative of composite tangent functions

Find the derivative of:

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

Solution:

  1. Let \( u(x) = 2x - 1 \), therefore \( \frac{du}{dx} = \frac{d}{dx}(2x - 1) = 2 \). Applying the rule: \[ \frac{d}{dx} f(x) = \sec^2 u \cdot \frac{du}{dx} = \sec^2(2x - 1) \cdot 2 = 2\sec^2(2x - 1) \]
  2. Let \( u(x) = \cos x \), therefore \( \frac{du}{dx} = \frac{d}{dx}\cos x = -\sin x \). Applying the rule: \[ \frac{d}{dx} g(x) = \sec^2 u \cdot \frac{du}{dx} = \sec^2(\cos x) \cdot (-\sin x) = -\sin x \sec^2(\cos x) \]
  3. Let \( u(x) = \frac{x-1}{x+2} \), therefore \( \frac{du}{dx} = \frac{3}{(x+2)^2} \) (using the quotient rule). Applying the rule: \[ \frac{d}{dx} h(x) = \sec^2 u \cdot \frac{du}{dx} = \sec^2\left(\frac{x-1}{x+2}\right) \cdot \frac{3}{(x+2)^2} = \frac{3\sec^2\left(\frac{x-1}{x+2}\right)}{(x+2)^2} \]

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