Question

y=e^x+e^x+e...

Answer 1

\[\LARGE y=e^{x+e^{x+e^{x+e}..}}\]

Answer 2

I dont know what to assume=y?

Answer 3

y= e^{x+e^x+...)
y=e^{x+y}

Answer 4

get that ^ ?

Answer 5

\[\LARGE logy=(x+y)loge?\]

Answer 6

and log e=1

Answer 7

we already have one y in the question,so we should name that to v suppose?

Answer 8

whats the problem in directly diff. log y =x+y ??

Answer 9

nothing :|

Answer 10

\[\LARGE \frac{dy}{dx} \times \frac{1}{y}=loge \frac{dy}{dx}+(x+y)\]

Answer 11

lol, log e was 1, right ?

Answer 12

yes,kar to dia :o

Answer 13

diferentiation of loge or actual value? O_O

Answer 14

right side, i get 1+dy/dx

Answer 15

\[\frac{dy}{dx} \frac{1}{y} = logeX(x+y) + (x+y)X loge\]
where X=differentition

Answer 16

let me write down steps
\( logy=(x+y)loge= x+y \\ (1/y )dy/dx= 1+dy/dx\)
get that ?

Answer 17

the value of log e=1 or derivative?

Answer 18

log e =1
derivative = 0

Answer 19

OH :o oky!