Question

$Prove space that space the space derivative space of space sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses comma space straight x greater than 0 space straight w. straight r. straight t. space square root of 1 minus straight x squared end root space is space equal space to space the space derivative space straight l subscript straight n space left parenthesis straight x squared right parenthesis space with space respect space to space straight x comma space left square bracket straight l subscript straight n left parenthesis straight x squared right parenthesis equals space log to the power of straight e. left parenthesis straight a squared right parenthesis right square bracket.$

Solution

Let space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses comma space straight u equals square root of 1 minus straight x squared end root Put space straight x equals cos space straight theta therefore space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 cos squared space straight theta minus 1 end fraction close parentheses equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator cos space 2 straight theta end fraction close parentheses equals sec to the power of negative 1 end exponent open parentheses sec space 2 straight theta close parentheses equals 2 straight theta therefore space straight y equals 2 space cos to the power of negative 1 end exponent straight x space space rightwards double arrow space dy over dx equals negative fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction Also space straight u equals square root of 1 minus straight x squared end root therefore space space space space space du over dx equals fraction numerator 1 over denominator 2 square root of 1 minus straight x squared end root end fraction cross times left parenthesis negative 2 straight x right parenthesis equals negative fraction numerator straight x over denominator square root of 1 minus straight x squared end root 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 negative fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction cross times fraction numerator square root of 1 minus straight x squared end root over denominator negative straight x end fraction space space space space space space space space space space space space space space space space space equals 2 over straight x equals 2 straight d over dx left parenthesis log subscript straight e space straight x right parenthesis equals straight d over dx left parenthesis 2 space log subscript straight e space straight x right parenthesis equals straight d over dx left parenthesis log subscript straight e space straight x squared right parenthesis equals straight d over dx left square bracket straight l subscript straight n left parenthesis straight x squared right parenthesis right square bracket

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