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

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:

This article belongs to the wider family of differentiation of trigonometric functions, 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