Fp1

http://mathsathawthorn.pbworks.com/f/FP1Jan07.pdf
In q6a we have to find the coordinates of the stationary point on the graph of f so so far ive worked out dy/dx which is (-2ylnx)/x and i know at a stationary point dy/dx=0 , but what do i do next?

Substitute in the value of y, and look for where the numerator is zero, i.e. one of the multiplicands is zero. Note that the domain is restricted to x > 0.

So if -2ylnx=0 How do i know the value of y and x?

You're told y as a function of x at the start, so sub that in, and you now have a product of functions of x. For that to equal zero, one of the multiplicands must be zero. Check each one to find what value of x - if any - will make it zero.

Step at a time please Okay so i have put they y value into x to get 0=-2^-lnx.lnx as the x cancels out?

EEK!!!! The x does not cancel.

dy/dx = -2x^{-ln x}{ln x}/x

= -2x^{-(ln x)-1}{ln x} = 0

So, either

ln x = 0, or

-2x^{-(ln x)-1} = 0

and recall that x>0, so can that second part ever be zero?

-2x^{-(ln x)-1}{ln x} = 0 where did the -1 come from? Im really sorry but idk :/

Rules of indices.

In essence,

x^a/ x = x^a / x^1 = x^(a-1)

EEK!!!! The x does not cancel.

dy/dx = -2x^{-ln x}{ln x}/x

= -2x^{-(ln x)-1}{ln x} = 0

So, either

ln x = 0, or

-2x^{-(ln x)-1} = 0

and recall that x>0, so can that second part ever be zero?

Rules of indices.

In essence,

x^a/ x = x^a / x^1 = x^(a-1)

I need help!