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EXERCISE 1.2 · Q106

Q.Differentiate the following w.r.t. xx: cot⁡−14−x−2x23x+2\cot^{-1}\dfrac{4-x-2x^2}{3x+2}

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Since the argument is positive (for suitable xx), cot⁡−1(4−x−2x23x+2)=tan⁡−1(3x+24−x−2x2)\cot^{-1}\left(\dfrac{4-x-2x^2}{3x+2}\right)=\tan^{-1}\left(\dfrac{3x+2}{4-x-2x^2}\right). Look for linear u,vu,v with u+v=3x+2u+v=3x+2 and 1−uv=4−x−2x21-uv=4-x-2x^2, i.e. uv=2x2+x−3uv=2x^2+x-3. Trying u=2x+3,v=x−1u=2x+3,v=x-1: u+v=(2x+3)+(x−1)=3x+2u+v=(2x+3)+(x-1)=3x+2 and uv=(2x+3)(x−1)=2x2+x−3uv=(2x+3)(x-1)=2x^2+x-3 — both match. So 3x+24−x−2x2=(2x+3)+(x−1)1−(2x+3)(x−1)=tan⁡[tan⁡−1(2x+3)+tan⁡−1(x−1)]\dfrac{3x+2}{4-x-2x^2}=\dfrac{(2x+3)+(x-1)}{1-(2x+3)(x-1)}=\tan[\tan^{-1}(2x+3)+\tan^{-1}(x-1)]. Hence $y=\tan^{-1}(2x+3)+ …

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