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EXERCISE 1.2 · Q105

Q.Differentiate the following w.r.t. xx: tan⁡−15−x6x2−5x−3\tan^{-1}\dfrac{5-x}{6x^2-5x-3}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Look for linear u,vu,v with u−v=5−xu-v=5-x and 1+uv=6x2−5x−31+uv=6x^2-5x-3 (matching tan⁡(P−Q)=tan⁡P−tan⁡Q1+tan⁡Ptan⁡Q\tan(P-Q)=\dfrac{\tan P-\tan Q}{1+\tan P\tan Q}), i.e. uv=6x2−5x−4uv=6x^2-5x-4. Trying u=2x+1,v=3x−4u=2x+1,v=3x-4: u−v=(2x+1)−(3x−4)=5−xu-v=(2x+1)-(3x-4)=5-x and uv=(2x+1)(3x−4)=6x2−5x−4uv=(2x+1)(3x-4)=6x^2-5x-4 — both match. So 5−x6x2−5x−3=(2x+1)−(3x−4)1+(2x+1)(3x−4)=tan⁡[tan⁡−1(2x+1)−tan⁡−1(3x−4)]\dfrac{5-x}{6x^2-5x-3}=\dfrac{(2x+1)-(3x-4)}{1+(2x+1)(3x-4)}=\tan[\tan^{-1}(2x+1)-\tan^{-1}(3x-4)]. Hence y=tan⁡−1(2x+1)−tan⁡−1(3x−4)y=\tan^{-1}(2x+1)-\tan^{-1}(3x-4). Differentia …

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