Mathematicsddxlog⁡exx−2x+23/4 is 

ddxlogexx2x+23/4 is 

  1. A

    1

  2. B

    x2+1x24

  3. C

    x21x24

  4. D

    exx21x24

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    Solution:

    let y=logexx2x+23/4
    =logex+logx2x+23/4y=x+34[log(x2)log(x+2)]
    On differentiating w.r.t. x, we get
    dydx=ddxx+34{log(x2)log(x+2)}=1+341x21x+2=1+3x24dydx=x21x24

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