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Miscellaneous Exercise · Q25

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

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This is a product of two factors, so we use the product rule. Differentiating each factor and combining gives 2x−tan⁡x−xsin⁡x+sin⁡xtan⁡x−xsec⁡2x+cos⁡x−sec⁡x2x - \tan x - x\sin x + \sin x\tan x - x\sec^2 x + \cos x - \sec x.

The expression (x+cos⁡x)(x−tan⁡x)(x + \cos x)(x - \tan x) is a product of two functions, so we differentiate it with the product rule: if y=u⋅vy = u\cdot v, then dydx=u′v+uv′\dfrac{dy}{dx} = u'v + uv'.

Step-by-step

  1. Name the two factors.

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

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

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

  1. Apply the product rule.

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

  1. Expand each product.

(1−sin⁡x)(x−tan⁡x)=x−tan⁡x−xsin⁡x+sin⁡xtan⁡x(1 - \sin x)(x - \tan x) = x - \tan x - x\sin x + \sin x\tan x

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

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