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Continuity And Differentiability

Question

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If y = (tan^-1 x)², then prove that (1 + x²)² y₂ + 2x(1 + x²)y_{1 = 2.}

Solution

space space space space space space space space straight y equals left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared 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 therefore space space space straight y subscript 1 equals 2 space tan to the power of negative 1 end exponent straight x. fraction numerator 1 over denominator 1 plus straight x squared end fraction rightwards double arrow space left parenthesis 1 plus straight x squared right parenthesis straight y subscript 1 equals 2 space tan to the power of negative 1 end exponent straight x rightwards double arrow space space left parenthesis 1 plus straight x squared right parenthesis squared straight y subscript 1 squared equals 4 open parentheses tan to the power of negative 1 end exponent straight x close parentheses squared rightwards double arrow space space left parenthesis 1 plus straight x squared right parenthesis squared straight y subscript 1 squared equals 4 straight y 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 square bracket because space of space left parenthesis 1 right parenthesis right square bracket Differentiation space both space sides space straight w. straight r. straight t. straight x comma space space space space space space space left parenthesis 1 plus straight x squared right parenthesis squared.2 straight y subscript 1 straight y subscript 2 plus 2 left parenthesis 1 plus straight x squared right parenthesis.2 straight x. straight y subscript 1 squared equals 4 straight y subscript 1 Divides space both space sides space ny space 2 straight y subscript 1 comma space we space get comma left parenthesis 1 plus straight x squared right parenthesis squared straight y subscript 2 plus 2 straight x left parenthesis 1 plus straight x squared right parenthesis straight y subscript 1 equals 2

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