Question

$Differentiate space colon space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x squared end root over denominator straight x end fraction close parentheses space straight w. straight r. straight t. space sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses.$

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Solution

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

Also

space space space space space space space straight u equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses Put space space straight x equals tan space straight theta therefore space space space space straight u equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 tan space straight theta over denominator 1 plus tan squared space straight theta end fraction close parentheses space space space space space space space space space space equals sin to the power of negative 1 end exponent left parenthesis sin space 2 straight theta right parenthesis space space space space space space space space space space equals 2 space straight theta therefore space space space space straight u equals 2 space tan to the power of negative 1 end exponent straight x therefore space du over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction

Now space dy over dx equals fraction numerator begin display style dy over dx end style over denominator begin display style du over dx end style end fraction equals fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction cross times fraction numerator 1 plus straight x squared over denominator 2 end fraction therefore space space space space space dy over dx equals 1 fourth

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