Let f(x) = x^x.Give a convicing argument showing that f'(2)~ 6.773.
The first way would be to take the derivative using logarithmic differentation.
$\displaystyle y=x^x \iff ln(y)=x\ln(x)$ Now taking the implicit derivative we get
$\displaystyle \frac{1}{y}y'=[1+ln(x)] \iff y'=y[1+\ln(x)]=x^x[1+\ln(x)]$
Now $\displaystyle f'(2)=2^2[1+ln(2)]=4+4\ln(2)=....$
Or you could approximate it by using the difference quotient
$\displaystyle f'(2) \approx =\frac{(2+h)^{2+h}-2^4}{h}$
Now use very small values of h like $\displaystyle h=10^{-6}$ or some other maybe smaller values.