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MISCELLANEOUS EXERCISE 1 (II) · Q236

Q.Differentiate tan⁡−1x1+6x2+cot⁡−11−10x27x\tan^{-1}\dfrac{x}{1+6x^2}+\cot^{-1}\dfrac{1-10x^2}{7x} w.r.t. x

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First term: using tan⁡−1A−tan⁡−1B=tan⁡−1A−B1+AB\tan^{-1}A-\tan^{-1}B=\tan^{-1}\dfrac{A-B}{1+AB} with A=3x, B=2xA=3x,\ B=2x: A−B1+AB=x1+6x2\dfrac{A-B}{1+AB}=\dfrac{x}{1+6x^2}. So tan⁡−1x1+6x2=tan⁡−1(3x)−tan⁡−1(2x)\tan^{-1}\dfrac{x}{1+6x^2}=\tan^{-1}(3x)-\tan^{-1}(2x).

Second term: using tan⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB\tan^{-1}A+\tan^{-1}B=\tan^{-1}\dfrac{A+B}{1-AB} with A=2x, B=5xA=2x,\ B=5x: A+B1−AB=7x1−10x2\dfrac{A+B}{1-AB}=\dfrac{7x}{1-10x^2}. So tan⁡−17x1−10x2=tan⁡−1(2x)+tan⁡−1(5x)\tan^{-1}\dfrac{7x}{1-10x^2}=\tan^{-1}(2x)+\tan^{-1}(5x); and since cot⁡−1z=tan⁡−1(1/z)\cot^{-1}z=\tan^{-1}(1/z) for z>0z>0, cot⁡−11−10x27x=tan⁡−17x1−10x2=tan⁡−1(2x)+tan⁡−1(5x)\cot^{-1}\dfrac{1-10x^2}{7x}=\tan^{-1}\dfrac{7x}{1-10x^2}=\tan^{-1}(2x)+\tan^{-1}(5x). …

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