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    For each of the following curves, find dy/dx and determine the exact x-coordinate of the stationary point:
    y= x^2/lnx

    i know that to differentiate this, i need to use the quotient rule:
    dy/dx = (lnx2x - x^2*1/x)/ (lnx)^2
    i made dy/dx equal to zero but i'm not sure how to solve this after that?
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    (Original post by elvss567)
    For each of the following curves, find dy/dx and determine the exact x-coordinate of the stationary point:
    y= x^2/lnx

    i know that to differentiate this, i need to use the quotient rule:
    dy/dx = (lnx2x - x^2*1/x)/ (lnx)^2
    i made dy/dx equal to zero but i'm not sure how to solve this after that?
    You have numerator = 0 so then just factor out the x
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    (Original post by RDKGames)
    You have numerator = 0 so then just factor out the x
    if i factor out x, i got:
    x(ln2 - x*1/x) / (lnx)^2

    the mark scheme says if i solve this, i must get x=e^1/2. how do i get this?
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    (Original post by elvss567)
    if i factor out x, i got:
    x(ln2 - x*1/x) / (lnx)^2

    the mark scheme says if i solve this, i must get x=e^1/2. how do i get this?
    Set it =0 then it’s obvious!
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    (Original post by RDKGames)
    Set it =0 then it’s obvious!
    I genuinely cant see it? could you elaborate please?
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    lnx\cdot 2x - x=0 -> x(2lnx-1)=0
    So, first solution is 0, and the next solution will be when you solve 2lnx-1=0
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    (Original post by elvss567)
    I genuinely cant see it? could you elaborate please?
    \displaystyle y'=\frac{2x \ln(x) - \frac{x^2}{x}}{\ln^2(x)} = \frac{x [2\ln(x)-1]}{\ln^2(x)}

    Set it =0 then clearly you have x(2\ln x -1) = 0... so either x=0 or 2 \ln(x)-1 =0. Former case is not possible.
 
 
 
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