Question

Differentiate the following functions by substitutions method : $cot to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses comma space straight x not equal to 0$

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Solution

Let space space space space space space straight y equals cot to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses Put space space space space space straight x equals tan space straight theta therefore space space space space space space space straight y equals cot to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses equals cot to the power of negative 1 end exponent open parentheses fraction numerator sec space straight theta minus 1 over denominator tan space straight theta end fraction close parentheses equals cot to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator 1 over denominator cos space straight theta end fraction end style minus 1 over denominator begin display style fraction numerator sin space straight theta over denominator cos space straight theta end fraction 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 1 minus cos space straight theta over denominator sin space straight theta end fraction close square brackets equals cot to the power of negative 1 end exponent open parentheses fraction numerator 2 sin squared begin display style straight theta over 2 end style over denominator 2 sin squared begin display style straight theta over 2 end style cos begin display style straight theta over 2 end style end fraction close parentheses equals cot to the power of negative 1 end exponent open parentheses tan straight theta over 2 close parentheses 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 cot open parentheses straight pi over 2 minus straight theta over 2 close parentheses close square brackets equals straight pi over 2 minus straight theta over 2 equals straight pi over 2 minus 1 half tan to the power of negative 1 end exponent straight x therefore space dy over dx equals 0 minus fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction equals negative fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction

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Continuity And Differentiability

Differentiate the following functions by substitutions method : $cot to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses comma space straight x not equal to 0$

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