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Worked Examples · Example 17

Q.Compute the derivative of tan⁡x\tan x.

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The derivative of tan⁡x\tan x is sec⁡2x\sec^2 x. This comes from rewriting tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x} and applying the quotient rule — a clean, standard result you'll use constantly in calculus.

The derivative of tan⁡x\tan x is one of those results that feels mysterious until you see where it comes from. Let's build it from first principles.

The Core Idea: Derivative at a Point

When we say "derivative of tan⁡x\tan x", we mean the function that gives the slope of the tangent line to y=tan⁡xy = \tan x at any point xx where it's defined. The definition is:

ddxtan⁡x=lim⁡h→0tan⁡(x+h)−tan⁡xh\frac{d}{dx} \tan x = \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h}

But working directly with tan⁡\tan is messy. A much cleaner path: rewrite tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x} and use the quotient rule. This works because we already know the derivatives of sin⁡x\sin x and cos⁡x\cos x, and the quotient rule handles the rest.

ddxtan⁡x=sec⁡2x\frac{d}{dx} \tan x = \sec^2 x

Let's prove it step by step.

  1. Rewrite tan⁡x\tan x in terms of sine and cosine.

tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}

This is valid wherever cos⁡x≠0\cos x \neq 0 (i.e., x≠π2+nπx \neq \frac{\pi}{2} + n\pi).

  1. Apply the quotient rule.

    For a function uv\frac{u}{v}, the derivative is u′v−uv′v2\frac{u'v - uv'}{v^2}. Here u=sin⁡xu = \sin x and v=cos⁡xv = \cos x.

    We know:

    • u′=cos⁡xu' = \cos x
    • v′=−sin⁡xv' = -\sin x

    So:

ddx(sin⁡xcos⁡x)=(cos⁡x)(cos⁡x)−(sin⁡x)(−sin⁡x)cos⁡2x\frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x}

  1. Simplify the numerator. The numerator becomes:

cos⁡2x+sin⁡2x\cos^2 x + \sin^2 x

This is exactly sin⁡2x+cos⁡2x\sin^2 x + \cos^2 x, which equals 11 (the fundamental Pythagorean identity). …

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