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Continuity And Differentiability

Question

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Differentiate the following function w.r.t.x: $left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared$

Solution

Let space space space space space straight y equals left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared therefore space dy over dx equals 2 tan to the power of negative 1 end exponent straight x. straight d over dx left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis equals 2 space tan to the power of negative 1 end exponent. fraction numerator 1 over denominator 1 plus straight x squared end fraction equals fraction numerator 2 space tan to the power of negative 1 end exponent over denominator 1 plus straight x squared end fraction

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