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EXERCISE 1.5 · Q211

Q.13x−5\dfrac{1}{3x-5}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Let y=13x−5=(3x−5)−1y=\dfrac{1}{3x-5}=(3x-5)^{-1}, of the form (ax+b)−1(ax+b)^{-1} with a=3, b=−5a=3,\,b=-5.

For y=x−1y=x^{-1} (i.e. a=1a=1), the pattern is yn=(−1)nn! x−n−1y_n=(-1)^n n!\,x^{-n-1} (from 4B). For a linear argument ax+bax+b, each differentiation contributes one extra factor of aa from the chain rule, so the pattern generalizes to: …

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