Question

y=x^x find y'

Answer 1

trickyy!

Answer 2

okay so this is what you do:
take ln on both sides:
lny = lnx^x

Answer 3

its a log... maybe

Answer 4

i got as far as lny=xlnx then Im lost

Answer 5

(..)^x * x'

Answer 6

now use implicit diff:
(1/y)(y') = [x*lnx]'

Answer 7

implicit differentiation

Answer 8

(1/y) y' = x*(1/x) + lnx*1
y' = y(1+lnx)

Answer 9

y' = x^x(1+lnx)

Answer 10

\[y=x^x\]
\[lny=lnx^x\]
\[lny=xlnx\]
\[\frac{y'}{y}=1*lnx+x*\frac{1}{x}\]
\[y'=y(lnx+1)=x^x(lnx+1)\]

Answer 11

\begin{eqnarray*}y' &=& (x^x)' \\
&=& (e^{x \log{x}})' \\
&=& e^{x \log{x}} (x \log{x})' \\
&=& x^x(\log{x} + 1).\end{eqnarray*}

Answer 12

\[y=f(x)^{g(x)}\]
\[lny=lnf(x)^{g(x)}\]
\[lny=g(x)lnf(x)\]
\[\frac{y'}{y}=g'(x)*lnf(x)+g(x)*\frac{f'(x)}{f(x)}\]
\[y'=y(g'(x)lnf(x)+g(x)\frac{f'(x)}{f(x)})=f^g*(g'*lnf+g*\frac{f'}{f})\]
f>0