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Continuity And Differentiability

Question

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Differentiate the following w.r.t.x: $cot to the power of negative 1 end exponent open parentheses square root of 1 plus straight x squared end root minus straight x close parentheses$

Solution

Let space space straight y equals cot to the power of negative 1 end exponent open parentheses square root of 1 plus straight x squared end root minus straight x close parentheses semicolon space Put space straight x equals tan space straight theta therefore space space space straight y equals cot to the power of negative 1 end exponent left square bracket square root of 1 plus tan squared straight theta end root minus tanθ right square bracket equals cot to the power of negative 1 end exponent left square bracket sec space straight theta minus tan space straight theta right square bracket space space space space space space space space space equals cot to the power of negative 1 end exponent open square brackets fraction numerator 1 over denominator cos space straight theta end fraction minus fraction numerator sin space straight theta over denominator cos space straight theta end fraction close square brackets equals cot to the power of negative 1 end exponent open square brackets fraction numerator 1 minus sin space straight theta over denominator cos space straight theta end fraction close square brackets space space space space space space space space space equals cot to the power of negative 1 end exponent open square brackets fraction numerator cos squared begin display style straight theta over 2 end style plus sin squared begin display style straight theta over 2 end style minus 2 sin begin display style straight theta over 2 end style cos begin display style straight theta over 2 end style over denominator cos squared straight theta over 2 minus sin squared straight theta over 2 end fraction close square brackets space space space space space space space space space equals cot to the power of negative 1 end exponent open square brackets fraction numerator open parentheses cos straight theta over 2 minus sin straight theta over 2 close parentheses squared over denominator open parentheses cos straight theta over 2 minus sin straight theta over 2 close parentheses open parentheses cos straight theta over 2 plus sin straight theta over 2 close parentheses end fraction close square brackets equals cot to the power of negative 1 end exponent open square brackets fraction numerator cos begin display style straight theta over 2 end style minus sin begin display style straight theta over 2 end style over denominator cos begin display style straight theta over 2 end style plus sin begin display style straight theta over 2 end style end fraction close square brackets space space space space space space space space space equals cot to the power of negative 1 end exponent open square brackets fraction numerator 1 minus tan straight theta over 2 over denominator 1 plus tan straight theta over 2 end fraction close square brackets equals cot to the power of negative 1 end exponent open square brackets cot open parentheses straight pi over 2 plus straight theta over 2 close parentheses close square brackets equals straight pi over 2 plus straight theta over 2 therefore space space space space straight y equals straight pi over 2 plus straight theta over 2 tan to the power of negative 1 end exponent straight x space space rightwards double arrow space dy over dx equals 0 plus 1 half. fraction numerator 1 over denominator 1 plus straight x squared end fraction equals fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction

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