Maths-
General
Easy

Question

The derivative of sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 x over denominator 1 plus x squared end fraction close parentheses with respect to cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 minus x squared over denominator 1 plus x squared end fraction close parentheses is

  1. -1
  2. 1
  3. 2
  4. 4

The correct answer is: 1


    Let p equals sin to the power of negative 1 end exponent invisible function application fraction numerator 2 x over denominator 1 plus x squared end fraction equals 2 tan to the power of negative 1 end exponent invisible function application x
    and q equals cos to the power of negative 1 end exponent invisible function application fraction numerator 1 minus x squared over denominator 1 plus x squared end fraction equals 2 tan to the power of negative 1 end exponent invisible function application x
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow dp over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction comma dq over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction end cell row cell therefore dp over dq equals fraction numerator dp over dx over denominator dq over dx end fraction equals fraction numerator fraction numerator open parentheses 1 plus straight x squared close parentheses over denominator 2 end fraction over denominator open parentheses 1 plus straight x squared close parentheses end fraction equals 1 end cell end table

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