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    Hi, I'm stuck on this question, it's just for 3 marks, it's probably quite easy.

    The Curve with the equation y=x^5/2.lnx/4, x>0 crosses the x axis at the point (4,0).

    The curve has a stationary point at R.

    Find the x-coordinate of R in exact form.

    (This is from a C3 Solomon Paper B)

    Thanks in advance
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    dy/dx = 0
    x^(3/2).(1 + (5/2)(ln[x/4])) = 0
    Therefore x^(3/2) = 0 or 1 + (5/2)(ln[x/4]) = 0

    so in exact form,
    (5/2)(ln[x/4]) = -1
    (ln[x/4]) = -2/5
    x/4 = e^(-2/5)
    x = 4e^(-2/5)

    I jus did that paper dis morning :P
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    Thank you
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    How did you get the dy/dx?

    When I did that I used the product rule and I got:

    x^3/2 + ln(x/4).(5/2)x^3/2
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    (Original post by the harry p)
    How did you get the dy/dx?

    When I did that I used the product rule and I got:

    x^3/2 + ln(x/4).(5/2)x^3/2
    What you have is the same as Featherflare it is just been factorised by Featherflare into a (slightly) neater form.

    (Original post by Featherflare)
    x^(3/2).(1 + (5/2)(ln[x/4])) = 0
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    (Original post by the harry p)
    How did you get the dy/dx?

    When I did that I used the product rule and I got:

    x^3/2 + ln(x/4).(5/2)x^3/2
    f(x) = x5/2.ln[¼x]
    u = x5/2
    v = ln[¼x]

    f'(x) = u.dv/dx + v.du/dx
    f'(x) = x5/2 x (¼)/(¼x) + ln[¼x] x [5/2]x3/2
    f'(x) = x5/2 x 1/x + ln[¼x] x [5/2]x3/2
    f'(x) = x3/2 + x3/2[5/2]ln[¼x]
    f'(x) = x3/2[1 + [5/2]ln[¼x]]
 
 
 
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