Question

$Differentiate space sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses space straight w. straight r. straight t. space sin to the power of negative 1 end exponent left parenthesis 3 straight x minus 4 straight x cubed right parenthesis.$

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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 Put space straight x equals cos space straight theta therefore space space 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 space space space space space space space space space equals sec to the power of negative 1 end exponent left parenthesis sec space 2 straight theta right parenthesis equals 2 straight theta equals 2 space cos to the power of negative 1 end exponent straight x therefore 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 space space space space space space space straight u equals space sin to the power of negative 1 end exponent left parenthesis 3 straight x minus 4 straight x cubed right parenthesis Put space straight x equals sin space straight theta therefore space space space straight u equals sin to the power of negative 1 end exponent left parenthesis 3 sin space straight theta minus 4 sin cubed space straight theta right parenthesis space space space space space space space space space equals sin to the power of negative 1 end exponent left parenthesis sin space 3 straight theta right parenthesis equals 3 straight theta equals 3 sin to the power of negative 1 end exponent therefore space du over dx equals fraction numerator 3 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 3 end fraction equals negative 2 over 3

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