Differentiation question

Steoob

hey im having some trouble differentiating (1-logeX)^2 {one minus log x to the base e squared}... i keep gettin -2/x but according to the answers in my book when i let dy/dx=0 x turns out to be e but i get -2.... any help pleease

seandoiler

y = (1 - ln[x])^2
dy/dx = 2 * (1 - ln[x]) * (-1/x) ............chain rule
so dy/dx = - 2 (1-ln[x]) / x
dy/dx=0 => -2 + 2 ln[x] = 0 => x=e

Steoob

o right... ya for some reason i put the -2 over x and the lnx over x and took that as ln(x/x) which is one hence ln1=0...

thanks very much this was annoying the **** out of me

analyse this

seandoiler wrote:

y = (1 - ln[x])^2
dy/dx = 2 * (1 - ln[x]) * (-1/x) ............chain rule
so dy/dx = - 2 (1-ln[x]) / x
dy/dx=0 => -2 + 2 ln[x] = 0 => x=e

You lost me there mate! Care to explain?:o

why is dy/dx=0?? and why does x=e?:o

seandoiler

well he said in his original post that he has to solve dy/dx =0, well he doesn't say it explicitly but rather implies it.....at any rate if you take dy/dx =0, you get -2 + 2 ln[x] =0, which gives 2 = 2 ln [x] => 1=ln[x] => e=x, this okay?

Steoob

i needed to find out where the there was a stationary point or something.. i cant really remember but i had to put dy/dx=0 to find x