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Continuity And Differentiability

Question

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

Solution

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

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