x, we have
1 dw
w dx
⋅
 =
(log )
log
( )
d
d
x
x
x
x
dx
dx
+
⋅
=
1
log
1
x
x
⋅x
+
⋅
i e dw
dx  = w (1 + log x)
= xx (1 + log x) (4)
From (1), (2), (3), (4), we have
log
log
x
y
x dy
y
dy
y
y
x
x
y dx
x
dx




+
+
+








 + xx (1 + log x) = 0
or
(x