Question

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

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

Let space space straight y 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 y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 space 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 equals 2 straight theta therefore space space space space straight y equals 2 tan to the power of negative 1 end exponent straight x space rightwards double arrow space dy over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction Also space straight u equals tan to the power of negative 1 end exponent therefore space space space space space du over dx equals fraction numerator 1 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 2 over denominator 1 plus straight x squared end fraction cross times fraction numerator 1 plus straight x squared over denominator 1 end fraction equals 2

For Daily Free Study Material Join wiredfaculty