Question

y=(lnx)/(1+lnx)

Answer 1

question?

Answer 2

To find the derivative of y please

Answer 3

I'm using the quotient rule. y'={ (1+lnx) * lnx' } - { lnx * (1+lnx)' }
y'= { (1+lnx) * (1/x) } - { lnx * (1+ 1/x) }
y' = (1+ lnx/x) - (lnx - lnx/x)
not sure if that's right.

Answer 4

i believe you're forgetting to divide by g^2(x)

Answer 5

(f(x)/g(x))' = (f'(x)g(x) - f(x)g'(x))/g^2(x)

Answer 6

I see what you're saying. Let me see if that fixes it.

Answer 7

The book has an answer of 1/x(1+lnx)^2. The answer that I get is 1+lnx / x(1+lnx)^2

Answer 8

the way you're formatting things is a bit confusing, you mean this right?

Answer 9

I see what I did wrong. Thanks for the help Bloopman!

y'=[{ (1+lnx) * lnx' } - { lnx * (1+lnx)' } ] /(1+lnx)^2
y'=[ { (1+lnx) * (1/x) } - { lnx * (1/x) }] /(1+lnx)^2
y' =[ (1+ lnx/x) - (lnx /x)] /(1+lnx)^2
y' = 1/ [x(1+lnx)^2]

Answer 10

yeah i probably would have caught that if the formatting wasn't so weird. sorry about that

Answer 11

No problem. First time on here so not sure how to do a proper format for easier reading but you guided me to the right answer. Thanks again!