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Miscellaneous Exercise · Q29

Q.Find the derivative of (x+sec⁡x)(x−tan⁡x)(x + \sec x)(x - \tan x).

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This is a product, so we use the product rule. After simplifying with tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1, the derivative is 2x−tan⁡x+xsec⁡xtan⁡x−xsec⁡2x+2sec⁡x−2sec⁡3x2x - \tan x + x\sec x\tan x - x\sec^2 x + 2\sec x - 2\sec^3 x.

The expression (x+sec⁡x)(x−tan⁡x)(x + \sec x)(x - \tan x) is a product of two functions, so the right tool is the product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'.

Step-by-step

  1. Name the two factors.

u=x+sec⁡x,v=x−tan⁡xu = x + \sec x, \qquad v = x - \tan x

  1. Differentiate each factor. Using ddx(sec⁡x)=sec⁡xtan⁡x\dfrac{d}{dx}(\sec x) = \sec x\tan x and ddx(tan⁡x)=sec⁡2x\dfrac{d}{dx}(\tan x) = \sec^2 x:

u′=1+sec⁡xtan⁡x,v′=1−sec⁡2xu' = 1 + \sec x\tan x, \qquad v' = 1 - \sec^2 x

  1. Apply the product rule.

dydx=(1+sec⁡xtan⁡x)(x−tan⁡x)+(x+sec⁡x)(1−sec⁡2x)\frac{dy}{dx} = (1 + \sec x\tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)

  1. Expand each product.

(1+sec⁡xtan⁡x)(x−tan⁡x)=x−tan⁡x+xsec⁡xtan⁡x−sec⁡xtan⁡2x(1 + \sec x\tan x)(x - \tan x) = x - \tan x + x\sec x\tan x - \sec x\tan^2 x

(x+sec⁡x)(1−sec⁡2x)=x−xsec⁡2x+sec⁡x−sec⁡3x(x + \sec x)(1 - \sec^2 x) = x - x\sec^2 x + \sec x - \sec^3 x

  1. Rewrite the term −sec⁡xtan⁡2x-\sec x\tan^2 x using tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1: −sec⁡xtan⁡2x=−sec⁡x(sec⁡2x−1)=−sec⁡3x+sec⁡x-\sec x\tan^2 x = -\sec x(\sec^2 x - 1) = -\sec^3 x + \sec x …

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