Question

Differentiate the following w.r.t.x: $cosec to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 2 straight x end fraction close parentheses$

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Solution

Let space space straight y equals cosec to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 2 straight x end fraction close parentheses semicolon space Put space straight x equals tan space straight theta therefore space space space space space straight y equals cosec to the power of negative 1 end exponent open parentheses fraction numerator 1 minus tan squared space straight theta over denominator 2 space tan space straight theta end fraction close parentheses equals cosec to the power of negative 1 end exponent left parenthesis cosec space 2 straight theta right parenthesis equals 2 straight theta therefore space space space space space straight y equals 2 space tan to the power of negative 1 end exponent straight x space space rightwards double arrow space dy over dx equals fraction numerator 1 over denominator 1 plus straight x squared end fraction

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