(x^lnx)' <= Why can't you apply the chain rule here?
-
If you can express x^lnx as the composition of two functions then you can apply the chain rule.
The difficulty is that there is no obvious way to express x^lnx as the composition of two functions.
So, you take ln of both sides of the equation y = x^lnx and get
lny = lnx lnx
so
lny = (lnx)^2
Now you do use the chain rule on both the left and right sides:
(1/y)y' = 2lnx 1/x, so
y' = 2y lnx /x
y' = 2(x^lnx)(lnx) / x
The difficulty is that there is no obvious way to express x^lnx as the composition of two functions.
So, you take ln of both sides of the equation y = x^lnx and get
lny = lnx lnx
so
lny = (lnx)^2
Now you do use the chain rule on both the left and right sides:
(1/y)y' = 2lnx 1/x, so
y' = 2y lnx /x
y' = 2(x^lnx)(lnx) / x