Question

trying to find derivative of y = x^(lnx), please help step by step?

Answer 1

i "ln"ed both sides so I have lny=ln(x^lnx)

Answer 2

Maybe its easier for you if you look at it this way:
\[ x^{\ln(x)} = e^{\ln(x)^2} \]

Answer 3

okay why is that, ythemanifold?

Answer 4

Change of basis: \[ x^{\ln(x)} = e^{\ln(x)\cdot \ln(x)} = e^{\ln(x)^2} \]

Answer 5

*base i meant :D

Answer 6

Then you can use the chain rule.

Answer 7

angelas answer is wrong, don't pay attention to it, no offense

Answer 8

haha ok, i'm going to ask my teacher to explain change of base but i am following you from there

Answer 9

ok then you need to apply the chain rule: \[ \frac{d}{dx} f(g(x)) = f^\prime(g(x))\cdot g^\prime(x) \] and the derivative of e^x: \[ \frac{d}{dx} e^x = e^x \]

Answer 10

That gives you \[ \frac{d}{dx} e^{\ln(x)^2} = (\ln(x)^2)^\prime\cdot e^{\ln(x)^2} \]

Answer 11

is there anyway to write that in terms of x and y? I'm confused

Answer 12

is f(g(x)) = y? and g(x) = x?

Answer 13

Well you can set \[ y=x^{\ln(x)} \] then the derivative is expressed as \[ \frac{dy}{dx} \]

Answer 14

i have to go to my next class but if you write out the steps I can check back in a couple of hours and interpret them. Thank you!

Answer 15

ok \[ (\ln(x)^2)^\prime = \frac{2\ln(x)}{x} \] Put that back into the expression above and you get \[ \frac{d}{dx} x^{\ln(x)} = \frac{2\ln(x)}{x} x^{\ln(x)} = 2\ln(x) x^{\ln(x)-1} \]

Answer 16

y=x^(ln(x))
ln(y)=ln(x^(ln(x))
ln(y)=ln(x)ln(x)
differentiate implicitly
(1/y)y'=(1/x)ln(x)+(1/x)ln(x)
(1/y)y'=(2/x)ln(x)
y'=(2/x)yln(x)
=(2yln(x))/x