Question

find dy/dx , y = ln(x)^x

Answer 1

y = (ln(x))^x

e^y = e^(x ln(x))

maybe is a good start?

Answer 2

\[e^{x\ln(x)}\] is it. then chain rule and product rule

Answer 3

am i missing a rule? because that does not really make sense to me to be honest

Answer 4

how do u know to do the e part?

Answer 5

the derivative of
\[e^x\] is
\[e^x\] so the derivative of
\[e^{x\ln(x)}=e^{x\ln(x)}\times \frac{d}{dx}[x\ln(x)]\] by the chain rule. all the work is taking the derivative of
\[x\ln(x)\] which is a product rule problem

Answer 6

so when you look at the problem how do you know to change ln(x)^x to e^xln(x)?

Answer 7

You have to perform the chain rule. For that, it'd be good to make a substitution:\[u=\ln(x),\]\[v=x,\]\[y=u^v.\]\[\frac{dy}{dx}=\frac{du^v}{du}\frac{du}{dx}+\frac{du^v}{dv}\frac{dv}{dx},\]\[\frac{du^v}{du}=u^{-1+v}v,\frac{du^v}{dv}=u^v\ln(u),\]\[\frac{du}{dx}=\frac{1}{x},\frac{dv}{dx}=1.\]Now plug in, in, in!

Answer 8

(and simplify)

Answer 9

so looks like you just did substitution right?

Answer 10

I substituted and performed both the chain and product rules.

Answer 11

ok i see. ty

Answer 12

i know it because we always change
\[f(x)^{g(x)}\] to
\[e^{g(x)\ln(f(x)})\]
that is pretty much the definition

Answer 13

whenever you get those variables in the arguement and the exponent ; its good to log or e it for simplicities sake

Answer 14

oh ok cool ty