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Continuity And Differentiability

Question

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Differentiate the following functions w.r.t.x: $cot to the power of negative 1 end exponent left parenthesis cosec space straight x plus cot space straight x right parenthesis$

Solution

Let space straight y equals cot to the power of negative 1 end exponent left parenthesis cosec space straight x plus cot space straight x right parenthesis equals cot to the power of negative 1 end exponent open square brackets fraction numerator 1 over denominator sin space straight x end fraction plus fraction numerator cos space straight x over denominator sin space straight x end fraction close square brackets space space space space space space space space space equals cot to the power of negative 1 end exponent open parentheses fraction numerator 1 plus cos space straight x over denominator sin space straight x end fraction close parentheses equals cot to the power of negative 1 end exponent open parentheses fraction numerator 2 cos squared begin display style straight x over 2 end style over denominator 2 sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style end fraction close parentheses space space space space space space space space space equals cot to the power of negative 1 end exponent open parentheses fraction numerator cos begin display style straight x over 2 end style over denominator sin begin display style straight x over 2 end style end fraction close parentheses equals cot to the power of negative 1 end exponent open square brackets cot straight x over 2 close square brackets equals straight x over 2 therefore space dy over dx equals 1 half

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