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Continuity And Differentiability

Question

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Differentiate the following w.r.t.x: $tan to the power of negative 1 end exponent left parenthesis square root of 1 plus straight x squared end root minus straight x right parenthesis$

Solution

Let space space straight y equals tan to the power of negative 1 end exponent left parenthesis square root of 1 plus straight x squared end root minus straight x right parenthesis space semicolon space Put space straight x equals tan space straight theta therefore space space space space straight y equals tan to the power of negative 1 end exponent left parenthesis square root of 1 plus tan squared straight theta end root minus tan space straight theta right parenthesis space space space space space space space space space space equals tan to the power of negative 1 end exponent left parenthesis sec space straight theta minus tan space straight theta right parenthesis equals tan to the power of negative 1 end exponent open parentheses 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 parentheses space space space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus sin space straight theta over denominator cos space straight theta end fraction close parentheses equals tan 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 begin display style straight theta over 2 end style minus sin squared begin display style straight theta over 2 end style end fraction close square brackets space space space space space space space space space space equals tan 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 tan to the power of negative 1 end exponent open square brackets fraction numerator cos straight theta over 2 minus sin straight theta over 2 over denominator cos straight theta over 2 plus sin straight theta over 2 end fraction close square brackets space space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 minus tan begin display style straight theta over 2 end style over denominator 1 plus tan straight theta over 2 end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 4 minus straight theta over 2 close parentheses close square brackets equals straight pi over 4 minus straight theta over 2 space space space space space space space space straight y equals straight pi over 4 minus 1 half tan to the power of negative 1 end exponent straight x space space rightwards double arrow space space dy over dx equals 0 minus fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction therefore space dy over dx equals negative fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction

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