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EXERCISE 1.2 · Q103

Q.Differentiate the following w.r.t. xx: cot⁡−1a2−6x25ax\cot^{-1}\dfrac{a^2-6x^2}{5ax}

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Since the argument is positive (for suitable x,ax,a), cot⁡−1(a2−6x25ax)=tan⁡−1(5axa2−6x2)\cot^{-1}\left(\dfrac{a^2-6x^2}{5ax}\right)=\tan^{-1}\left(\dfrac{5ax}{a^2-6x^2}\right). Look for A,BA,B with A+B=5xaA+B=\dfrac{5x}{a} and AB=6x2a2AB=\dfrac{6x^2}{a^2} (dividing numerator and denominator by a2a^2); trying A=2xa,B=3xaA=\dfrac{2x}{a},B=\dfrac{3x}{a} gives A+B=5xaA+B=\dfrac{5x}{a} and AB=6x2a2AB=\dfrac{6x^2}{a^2} — both match. So y=tan⁡−1(2xa)+tan⁡−1(3xa)y=\tan^{-1}\left(\dfrac{2x}{a}\right)+\tan^{-1}\left(\dfrac{3x}{a}\right). …

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