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# Another C3 question watch

1. 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)

2. 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

3. Thank you
4. 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
5. (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
6. (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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