The Student Room Group
Reply 1
jenh1
curve with equation y=x^(5/2)ln(x/4) crosses x-axis at (4,0)

the curve has a stationary point at R

find the x coordinate of R in exact form?


y = x5/2ln(x/4)
y = (x5)1/2ln[x/4]
y = √x5ln[x/4]
dy/dx = d[√x5]/dx.ln[x/4] - d[ln[x/4]]/dx.√x5
....... = 5/2x3/2ln[x/4] - 1/x√x5
....... = 5/2√x³ln[x/4] - 1/x√x5
....... = √x³(5/2ln[x/4] - 1)

sub (x = 4) into dy/dx ...
....... = √4³(5/2ln[4/4] - 1)
....... = √64(5/2ln1 - 1)
....... = -8
Reply 2
jenh1
curve with equation y=x^(5/2)ln(x/4) crosses x-axis at (4,0)

the curve has a stationary point at R

find the x coordinate of R in exact form?


y = x5/2ln(x/4)
y = (x5)1/2ln[x/4]
y = √x5ln[x/4]
dy/dx = d[√x5]/dx.ln[x/4] - d[ln[x/4]]/dx.√x5
....... = 5/2x3/2ln[x/4] - 1/x√x5
....... = 5/2√x³ln[x/4] - 1/x√x5
....... = √x³(5/2ln[x/4] - 1)

sub (dy/dx = 0) ...
....... => √x³(5/2ln[x/4] - 1) = 0

x = 0? not sure
Reply 3
jenh1
curve with equation y=x^(5/2)ln(x/4) crosses x-axis at (4,0)

the curve has a stationary point at R

find the x coordinate of R in exact form?


Here is my solution.
Reply 4
To find dy/dx:

Using product rule:

y = x^(5/2)ln(x/4)
y = uv

where u = x^(5/2)
and v = ln(x/4)

du/dx = (5/2)x^(3/2)

and find dv/dx using the chain rule:

dv/dx = dv/dw x dw/dx where w = x/4

dv/dw = 1/w
dw/dx = 1/4

dv/dx = 4/x mulitplied by 1/4
= 1/x

Back to product rule:

dy/dx = uv' + vu'
= x^(5/2).(1/x) + (ln(x/4)).(5/2)x^(3/2)
= x^(3/2) + ln(x/4).(5/2)x^(3/2)

Then you say dy/dx = 0 and go from there.

I'm pretty sure it's possible from there but I've forgotten how to do logs. :biggrin:
Reply 5
Joe_87


Then you say dy/dx = 0 and go from there.

I'm pretty sure it's possible from there but I've forgotten how to do logs. :biggrin:


The question asks for the exact value of the x co-ordinate of R. Steve has answered the question.
Reply 6
carol05
The question asks for the exact value of the x co-ordinate of R. Steve has answered the question.


Yes I do know. Well done to steve. He happened to finish his post before I did, otherwise I wouldn't have bothered with my answer, which I am fully aware is incomplete.