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Math Help - Natural Exponential Function#6

  1. #1
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    Natural Exponential Function#6

    Find an equation for the tangent line to the curve y=(e^x)/(1+lnx) at the point where x=1.

    Can someone check my work please?

    y=e^x/(1+lnx)

    y'=[e^x(1+lnx)-e^x(1/x)]/(1+lnx)^2

    f'(1)=e(0)/1
    =0

    y=mx+b
    e=0(1)+b
    e=b

    therefore the equation of the tangent is
    y=e

    Is this correct? thank you in advance!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by skeske1234 View Post
    Find an equation for the tangent line to the curve y=(e^x)/(1+lnx) at the point where x=1.

    Can someone check my work please?

    y=e^x/(1+lnx)

    y'=[e^x(1+lnx)-e^x(1/x)]/(1+lnx)^2

    f'(1)=e(0)/1
    =0

    y=mx+b
    e=0(1)+b
    e=b

    therefore the equation of the tangent is
    y=e

    Is this correct? thank you in advance!

    it is correct
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