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Continuity And Differentiability

Question

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Differentiate the following w.r.t.x:
$open parentheses 1 over straight x close parentheses to the power of straight x$

Solution

Let space straight y equals open parentheses 1 over straight x close parentheses to the power of straight x therefore space log space straight y equals log open parentheses 1 over straight x close parentheses to the power of straight x space space space space space space space space space space space rightwards double arrow space log space straight y equals straight x space log space 1 over straight x rightwards double arrow space log space straight y equals straight x left parenthesis log space 1 minus log space straight x right parenthesis space rightwards double arrow log space straight y equals negative left square bracket straight x space log space straight x right square bracket space space space space left square bracket because space log space 1 equals 0 right square bracket Differentiating space both space sides space straight w. straight r. straight t space space straight x. 1 over straight y dy over dx equals negative open square brackets straight x.1 over straight x plus log space straight x.1 close square brackets rightwards double arrow space dy over dx equals negative straight y left square bracket 1 plus log space straight x right square bracket equals negative open parentheses 1 over straight x close parentheses to the power of straight x left square bracket 1 plus log space straight x right square bracket.

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