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Continuity And Differentiability

Question

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$Prove space that space cot to the power of negative 1 end exponent straight x space plus space cot minus 1 space 1 over straight x space is space constant.$

Solution

Let space space space space space straight y equals cot to the power of negative 1 end exponent straight x plus cot to the power of negative 1 end exponent 1 over straight x therefore space dy over dx equals negative fraction numerator 1 over denominator 1 plus straight x squared end fraction minus fraction numerator 1 over denominator 1 plus begin display style 1 over straight x squared end style end fraction. straight d over dx open parentheses 1 over straight x close parentheses equals negative fraction numerator 1 over denominator 1 plus straight x squared end fraction minus fraction numerator straight x squared over denominator 1 plus straight x squared end fraction cross times open parentheses negative 1 over straight x squared close parentheses space space space space space space space space space space space space space equals negative fraction numerator 1 over denominator 1 plus straight x squared end fraction plus fraction numerator 1 over denominator 1 plus straight x squared end fraction therefore space straight d over dx left parenthesis straight y right parenthesis equals 0 space rightwards double arrow space straight y equals constant therefore space cot to the power of negative 1 end exponent straight x plus cot to the power of negative 1 end exponent 1 over straight x space space is space constant

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