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core 3 differentiation

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?
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
Reply 2
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?
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!
Reply 4
Original post by RDKGames
Set it =0 then it’s obvious!


I genuinely cant see it? could you elaborate please?
lnx2xx=0lnx\cdot 2x - x=0 -> x(2lnx1)=0x(2lnx-1)=0
So, first solution is 0, and the next solution will be when you solve 2lnx1=02lnx-1=0
Original post by elvss567
I genuinely cant see it? could you elaborate please?


y=2xln(x)x2xln2(x)=x[2ln(x)1]ln2(x)\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(2lnx1)=0x(2\ln x -1) = 0... so either x=0x=0 or 2ln(x)1=02 \ln(x)-1 =0. Former case is not possible.

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