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    can't seem to figure out q:
    Find the equation of the tangent and the normal to the curve y=e^x(lnx) at the point where x=1
    I worked the first derivative to be dy/dx=e^x(lnx) + e^x/x
    when i sub x=1 to find gradient of tangent i get e but from this point i go wrong !
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    (Original post by Wisteria_xo)
    can't seem to figure out q:
    Find the equation of the tangent and the normal to the curve y=e^x(lnx) at the point where x=1
    I worked the first derivative to be dy/dx=e^x(lnx) + e^x/x
    when i sub x=1 to find gradient of tangent i get e but from this point i go wrong !
    Correct so far.

    But how did you go wrong after that? What did you try?
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    (Original post by SeanFM)
    Correct so far.

    But how did you go wrong after that? What did you try?
    Gradient of tangent = e
    Tried to find the equation of the tangent, x=1 y=0 and c=-2 therefore i got y=ex-e
    however the mark scheme from the book said y=ex is eq of tangent.

    Finding normal I got gradient = -1/e and c=1/e therefore i got y=-1/ex +1/e
    which simplifies to ey=-x+1
    again the book gives a different answer ey+x = 1+e^2
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    (Original post by Wisteria_xo)
    Gradient of tangent = e
    Tried to find the equation of the tangent, x=1 y=0 and c=-2 therefore i got y=ex-e
    however the mark scheme from the book said y=ex is eq of tangent.

    Finding normal I got gradient = -1/e and c=1/e therefore i got y=-1/ex +1/e
    which simplifies to ey=-x+1
    again the book gives a different answer ey+x = 1+e^2
    :hmmmm: is there any chance that  f(x) = e^{xlnx} ? You can already see that the equation of the tangent is wrong as f(1) is not 0. It is always possible that the textbook is wrong.
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    (Original post by SeanFM)
    :hmmmm: is there any chance that  f(x) = e^{xlnx} ? You can already see that the equation of the tangent is wrong as f(1) is not 0. It is always possible that the textbook is wrong.
    no the f(x) is written correctly aha i suppose our textbook is famous for its mistakes !
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    (Original post by Wisteria_xo)
    no the f(x) is written correctly aha i suppose our textbook is famous for its mistakes !
    f(x)=e^x\ln(x) \longrightarrow f(1)=0 \Rightarrow f'(x)=\frac{e^x}{x}(x\ln{x} + 1) \longrightarrow f'(1)=e

    So tangent is y=e(x-1) and normal is y=-\frac{1}{e}(x-1)

    If this is not what the book shows, then it is wrong.
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    (Original post by RDKGames)
    f(x)=e^x\ln(x) \longrightarrow f(1)=0 \Rightarrow f'(x)=\frac{e^x}{x}(x\ln{x} + 1) \longrightarrow f'(1)=e

    So tangent is y=e(x-1) and normal is y=-\frac{1}{e}(x-1)

    If this is not what the book shows, then it is wrong.
    I got exactly the same ! I guess it must be wrong then, thank you so much!
 
 
 
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