Question

find dy/dx of y=lnx-1/e^x+1

Answer 1

is it:
\[y=\ln(x) - 1/(e ^{x}+1)\]
?

Answer 2

\[y=\frac{lnx-1}{e^x+1} ?\]

Answer 3

yep

Answer 4

well the way he has it is what you have inink

Answer 5

yep to which one?

Answer 6

say first or second one

Answer 7

second.
thanks guys

Answer 8

\[y'=\frac{(lnx-1)'(e^x+1)-(e^x+1)'(lnx-1)}{(e^x+1)^2}\]

Answer 9

(lnx-1)'=1/x-0=1/x

(e^x+1)=e^x+0=e^x

Answer 10

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

Answer 11

\[y'=(1/x *(e ^{x}+1) -e ^{x}(lnx-1))/(e ^{x}+1)^{2}\]
see myininaya's ...
simplify if needed

Answer 12

oooo...see i was confused about what to do with the [e ^{x}. thought i had to do somethin tricksy lol. thanks a lot =)

Answer 13

\[y'=\frac{x}{x}*\frac{\frac{1}{x}*(e^x+1)-e^x(lnx-1)}{(e^x+1)^2}=\frac{(e^x+1)-xe^x(lnx-1)}{x(e^x+1)^2}\]