Question

what is the derivative of x^2x or for that matter x^nx?

Answer 1

shouldnt this just come on to be 2x(x) -> 2x^2?
x^nx -> nx x ?

Answer 2

is this
\[ x^{nx} \]
?

Answer 3

correct

Answer 4

I think it's a bit tricky
y= x^(nx)
take the natural log (ln) of both sides
ln(y)= ln ( x^(nx))
ln(y)= nx* ln(x)
now take the derivative
1/y dy/dx = nx (d/dx ln(x)) + ln(x)* d/dx (nx)
replace 1/y with x^(-nx) and complete the derivative on the right side
x^(-nx) dy/dx = nx*(1/x) + n*ln(x)
dy/dx= n(1+ln(x))x^(nx)

Answer 5

what if i dont have this set equal to y just the plain derivative of x^nx or is that even possible?

Answer 6

the stuff on the right hand side is the "plain derivative"
setting it equal to y just makes it easier to type up

Answer 7

that works then, because what i am asked here is to evaluate a limit with working oh L'Hopital's rule
\[\lim_{x \rightarrow 0^-} x^{2x}\]

Answer 8

actually sorry its \[\lim_{x \rightarrow 0^+}\]

Answer 9

the answer for this problem is 1 as the limit approaches 0+

Answer 10

also say for example, i have another question - taking the derivative of \[(\tan x)^{\cos x}\]

Answer 11

\[(1+\frac{ 1 }{ x })^{\ln x}\]

Answer 12

L'Hopital's rule requires \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \)
to use it with \(x^{2x} \)
I would rewrite it as
\[\lim_{x \rightarrow 0+}x^{2x}= e^{\lim_{x \rightarrow 0+} 2x \ln(x)}\]
now rewrite x lnx as
\[ \frac{\ln x}{x^{-1}} \]
This is now in a form where you can apply L'Hopital's rule
d/dx ln x = 1/x
d/dx x^-1 = -x^2
the ratio is -x
and
\[ \lim_{x->0+} -x =0\]
we get
\[ e^{\lim_{x \rightarrow 0+} 2x \ln(x)}= e^0=1\]

Answer 13

Thank you!