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EXERCISE 1.2 · Q99

Q.Differentiate the following w.r.t. xx: cot⁡−11+35x22x\cot^{-1}\dfrac{1+35x^2}{2x}

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Since the argument is positive, cot⁡−1(1+35x22x)=tan⁡−1(2x1+35x2)\cot^{-1}\left(\dfrac{1+35x^2}{2x}\right)=\tan^{-1}\left(\dfrac{2x}{1+35x^2}\right). Look for A,BA,B with A−B=2xA-B=2x and 1+AB=1+35x21+AB=1+35x^2, i.e. AB=35x2AB=35x^2; trying A=7x,B=5xA=7x,B=5x gives A−B=2xA-B=2x and AB=35x2AB=35x^2 — both match. So 2x1+35x2=7x−5x1+(7x)(5x)=tan⁡[tan⁡−1(7x)−tan⁡−1(5x)]\dfrac{2x}{1+35x^2}=\dfrac{7x-5x}{1+(7x)(5x)}=\tan[\tan^{-1}(7x)-\tan^{-1}(5x)]. Hence $y=\tan …

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