log question (1 Viewer)

[lourai*87]

Sorry...this is probably really simple but ive done it over and over and it just wont work for me:

Find the equation of the tangent to the curve y = log_ex at the point (2, log_e2)

when i eventually differentiate to find the gradient..do i sub 0 for x or 2 for x ? But either way i still cant get it..thanks

 

[Dreamerish*~]

Sorry...this is probably really simple but ive done it over and over and it just wont work for me:



when i eventually differentiate to find the gradient..do i sub 0 for x or 2 for x ? But either way i still cant get it..thanks

you sub 2 for x

 

[Jago]

you sub in the x value for x. in the point (2, log_e2) it's it first value: 2.

 

[lourai*87]

lol....after the god-knows-how-many times over the last few days and i get it first go today...

thanks though!! This wont be the last question i post dont worry

 

[lourai*87]

See..here we go again.

Find the stationary point on the curve y = lnx/x and determine its nature.

I know how to do this normally, but the answer i dont understand, and i end up with some messy answer. Thanks

 

[withoutaface]

dy/dx= (1/x*x-lnx)/x^2=(1-lnx)/x^2
The stationary point is where 1-lnx=0 :. x=e

 

[lourai*87]

i did get to that point surprisingly, i wasnt sure if i differentiated properly..but how come x=e -- its too late at night for this stuff

 

[Slidey]

Find the stationary point on the curve y = lnx/x and determine its nature.
y'=1/x^2-lnx/x^2=0
1-lnx=0
lnx=1
x=e
Testing e^-1 and e^2 (either side of it):
x=e^-1, y=-1/e
x=e^2, y=2/e
Thus (x,y)=(e,1/e) is a maximum value

 

[lourai*87]

thank you guys so much!! Im such an idiot, i just couldnt work out why x = e. I think i get lazy and give up too easily

You guys are do helpful...and yes you will be sick of me soon enough, but hey