# Derivative of Tan 2x - Formula, Proof, and Examples

TL;DR

The derivative of tan 2x is \( \dfrac{d}{dx}(\tan 2x) = 2\sec^2 2x \). This article proves that result three ways — chain rule, first principle, and quotient rule — works through examples, and clears up the most common confusion: \( \tan 2x \) (double angle) versus \( \tan^2 x \) (tangent squared), which have different derivatives.

## What Is the Derivative of Tan 2x?

The **derivative of tan 2x** with respect to x is:

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

Here \( \sec \) is the secant function, the reciprocal of cosine: \( \sec \theta = \dfrac{1}{\cos \theta} \). The result comes from the standard derivative \( \dfrac{d}{dx}(\tan u) = \sec^2 u \) combined with the chain rule, because \( \tan 2x \) is a composite function — a tangent wrapped around an inner function \( 2x \).

Before any proof, name the pieces:

- **Outer function:** \( \tan u \), whose derivative is \( \sec^2 u \).
- **Inner function:** \( u=2x \), whose derivative is \( 2 \).

This article belongs to the wider family of [differentiation of trigonometric functions](/content/math/trigonometry/differentiation-of-trigonometric-functions/index.html), which share the same chain-rule pattern.

## How Do You Prove the Derivative of Tan 2x?

Three proofs land on the same answer. Seeing more than one is the point — the chain rule is fastest, but the first principle shows _why_ it works.

### Proof 1 - Chain Rule

The chain rule says \( \dfrac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). With \( f(u)=\tan u \) and \( g(x)=2x \):

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

### Proof 2 - Quotient Rule

Write \( \tan 2x = \dfrac{\sin 2x}{\cos 2x} \) and apply the quotient rule \( \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2} \):

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

### Proof 3 - First Principle

The first principle uses the limit definition \( \dfrac{d}{dx} = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} \). With \( f(x)=\tan 2x \):

\[ f'(x)=\lim_{h \to 0} \dfrac{\tan(2x + 2h) - \tan(2x)}{h} \]

Using the identity \( \tan A - \tan B = \dfrac{\sin(A - B)}{\cos A \cos B} \):

\[ f'(x)=\lim_{h \to 0} \dfrac{1}{h} \cdot \dfrac{\sin(2h)}{\cos(2x + 2h)\cos(2x)} \]

Split off the standard limit: \( \lim_{h \to 0}\dfrac{\sin 2h}{2h} = 1 \):

\[ f'(x)=2\sec^2(2x) \]

All three agree: \( \dfrac{d}{dx}(\tan 2x) = 2\sec^2 2x \)

## Examples of the Derivative of Tan 2x

### Example 1: Differentiate \( y=\tan 2x \) and find the slope at \( x=0 \).

\[ \dfrac{dy}{dx} = 2\sec^2 2x \]

At \( x=0 \): \( \sec 0 = 1 \), so \( \dfrac{dy}{dx} = 2(1)^2 = 2 \).

**Final answer:** slope = 2.

### Example 2: Differentiate \( y=\tan 2x \) versus \( y=\tan^2 x \).

Differentiating \( \tan 2x \) gives \( \dfrac{d}{dx}(\tan 2x) = 2\sec^2 2x \) and for \( \tan^2 x \): \( \dfrac{d}{dx}(\tan x)^2 = 2\tan x \cdot \sec^2 x \).

**Final answer:** different functions, different derivatives.

### Example 3: Differentiate \( y=\tan(2x+1) \).

\[ \dfrac{dy}{dx} = 2\sec^2(2x+1) \]

**Final answer:** 2sec²(2x+1).

### Example 4: Differentiate \( y=\tan(\tan 2x) \).

\[ \dfrac{dy}{dx} = \sec^2(\tan 2x) \cdot 2\sec^2 2x \]

**Final answer:** 2sec²(tan 2x).

### Example 5: Differentiate \( y=\tan 2x + \sec 2x \).

\[ \dfrac{dy}{dx} = 2\sec^2 2x + 2\sec 2x \tan 2x \]

**Final answer:** 2sec² 2x + 2sec 2x tan 2x.

### Example 6: Find the second derivative of \( y=\tan 2x \).

\[ y'' = 8\sec^2 2x \tan 2x \]

**Final answer:** y'' = 8sec² 2x tan 2x.

## Why This Derivative Matters - "Recording a doubled rate of change"

The reason this derivative earns attention is that the factor of 2 is not cosmetic. Differentiation measures rate of change, and \( \tan 2x \) varies twice as fast as \( \tan x \) because its angle advances twice as quickly.

## Common Mistakes With the Derivative of Tan 2x

### Mistake 1: Forgetting the chain-rule factor of 2

**Correct way:** Multiply by the inner derivative: \( 2\sec^2 2x \).

### Mistake 2: Confusing \( \tan 2x \) with \( \tan^2 x \)

**Correct way:** Differentiate them correctly to avoid confusion.

### Mistake 3: Wrong derivative for the secant term

**Correct way:** Use the formula for differentiating secant: \( \dfrac{d}{dx}(\sec u) = \sec u \tan u \cdot u' \).

## Key Takeaways

- The **derivative of tan 2x** is \( 2\sec^2 2x \).
- It can be proved by the chain rule, the quotient rule, or the first principle — all agree.
- In general, \( \dfrac{d}{dx}(\tan nx) = n\sec^2 nx \).