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Continuity And Differentiability

Question

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Differentiate the following w.r.t.x: $square root of fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction end root$

Solution

Let space space space space space straight y equals square root of fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction end root therefore dy over dx equals fraction numerator 1 over denominator square root of fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction end root end fraction. straight d over dx open parentheses fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction close parentheses equals fraction numerator square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root end fraction. fraction numerator left parenthesis 1 minus straight x right parenthesis.1 minus left parenthesis 1 plus straight x right parenthesis left parenthesis negative 1 right parenthesis over denominator left parenthesis 1 minus straight x right parenthesis squared end fraction space space space space space space space space space space space space equals fraction numerator square root of 1 minus straight x end root over denominator square root of fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction end root end fraction. fraction numerator 1 minus straight x plus 1 plus straight x over denominator left parenthesis 1 minus straight x right parenthesis squared end fraction equals fraction numerator square root of 1 minus straight x end root over denominator square root of 1 plus straight x end root end fraction. fraction numerator 2 over denominator left parenthesis 1 minus straight x right parenthesis squared end fraction equals fraction numerator 1 over denominator square root of 1 plus straight x end root left parenthesis 1 minus straight x right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction

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