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Continuity And Differentiability

Question

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Differentiate the following w.r.t.x: $log open parentheses straight x plus square root of 1 plus straight x squared end root close parentheses$

Solution

Let space straight y equals log open parentheses straight x plus square root of 1 plus straight x squared end root close parentheses therefore space dy over dx equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction. straight d over dx open parentheses straight x plus square root of 1 plus straight x squared end root close parentheses equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction. open parentheses 1 plus fraction numerator 2 straight x over denominator 2 square root of 1 plus straight x squared end root end fraction close parentheses space space space space space space space space space space space space space equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction. open parentheses 1 plus fraction numerator straight x over denominator 2 square root of 1 plus straight x squared end root end fraction close parentheses equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction cross times fraction numerator square root of 1 plus straight x squared end root plus straight x over denominator square root of 1 plus straight x squared end root end fraction therefore space dy over dx equals fraction numerator 1 over denominator square root of 1 plus straight x squared end root end fraction

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