Question

$Differentiate space open parentheses log subscript cos space straight x end subscript space sin space straight x close parentheses open parentheses log subscript sin space straight x end subscript space cos space straight x close parentheses plus cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses space straight w. straight r. straight t. straight x space at space straight x equals straight pi over 4.$

Solution

Let space space space straight y equals open parentheses log subscript cos space straight x end subscript space sin space straight x close parentheses open parentheses log subscript sin space straight x end subscript space cos space straight x close parentheses plus cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses Put space space space straight u equals open parentheses log subscript cos space straight x end subscript space sin space straight x close parentheses open parentheses log subscript sin space straight x end subscript space cos space straight x close parentheses comma space straight v equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses therefore space space space space space space straight y equals straight u plus straight v therefore space dy over dx equals du over dx plus dv over dx space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space straight u equals open parentheses log subscript cos space straight x end subscript space sin space straight x close parentheses open parentheses log subscript sin space straight x end subscript space cos space straight x close parentheses equals fraction numerator log space sin space straight x over denominator log space cos space straight x end fraction cross times fraction numerator log space cos space straight x over denominator log space sin space straight x end fraction equals 1 therefore space du over dx equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis space space straight v equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses Put space straight x equals tan space straight theta therefore space space space straight v equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus tan squared space straight theta over denominator 1 plus tan squared space straight theta end fraction close parentheses equals space space straight v equals cos to the power of negative 1 end exponent open parentheses cos space 2 straight theta close parentheses equals 2 straight theta equals 2 space tan to the power of negative 1 end exponent straight x therefore space dv over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis

From (1), (2), (3), we get,

space space space space space space dy over dx equals 0 plus fraction numerator 2 over denominator 1 plus straight x squared end fraction At space straight x equals straight pi over 4 comma space dy over dx equals fraction numerator 2 over denominator 1 plus open parentheses begin display style straight pi over 4 end style close parentheses squared end fraction equals fraction numerator 2 over denominator 1 plus begin display style straight pi squared over 16 end style end fraction equals fraction numerator 32 over denominator straight pi squared plus 16 end fraction

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