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Continuity And Differentiability

Question
CBSEENMA12035173

Differentiate the following functions w.r.t.x: 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

Solution
Let space space space space space straight y 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 square brackets 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 square brackets
space space space space space space space space space space space space space equals cot to the power of negative 1 end exponent open square brackets 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 square brackets 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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