y
y
.
,
,
∞
=
log
e
(
x
+
log
e
(
x
+
...
.
.
)
)
,
then
d
y
d
x
a
t
(
x
=
e
2
−
2
,
y
√
2
)
is
A
log
(
e
2
)
2
√
2
(
e
2
−
1
)
Mathbot Says...
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.
y
y
.
,
,
∞
=
log
e
(
x
+
log
e
(
x
+
...
.
.
)
)
,
then
d
y
d
x
a
t
(
x
=
e
2
−
2
,
y
√
2
)
is
A
log
(
e
2
)
2
√
2
(
e
2
−
1
)
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.