differentiate x^(ln(x))?

Answer 1

IIf you had trouble following Pascal's correct result, here is an alternative derivation:
Let y = x^(ln(x))

ln(y) = ln^2(x)
y' / y = 2 / x * ln (x)
y' = y * 2 / x * ln (x)
y' = x^(ln(x)) * 2 * ln(x) / x

I verified by plugging answer into Wolfram integrator and getting result x^(log(x)) (see link below).

Answer 2

Differentiate X Ln X

Answer 3

well the derivative of ln(x) is 1/x. so you take and put your ln(x) down in front. This is all part of the chain rule. So far you have ln(x) then you take the derivative of the whole thing, which when you do x^ln you get e. so now you have ln(x)*e^(lnx))^2 and then you pull your 2 down also and you get as your final answer: (2*ln(x)*e^(ln(x))^2)/2

Answer 4

y=e^{ln (x^{ln x})}
=e^{ln(x) ln(x)}
=e^{ln(x)^2}
chain rule:
y'=e{ln(x) ^2} 2 ln(x) 1/x
=[x^{ln(x)} 2 ln(x) ]/x
=2x^{ln(x) -1} ln(x)

Answer 5

x^(ln x) = e^(ln² x)
d(e^(ln² x))/dx = x^(ln x) * d(ln² x)/dx
d(ln² x)/dx = 2 ln x / x

d(e^(ln² x))/dx = 2 x^(ln x - 1) ln x

Answer 6

(2*ln(x)*e^(ln(x))^2)/x

Answer 7

y=x**lnx

lnx+(x**lnx)/(xlnx)