Question

Derivata funcției f:R-> R,

f(x)= |x|^|x|, daca x diferit de 0
1, daca x=0

|x|^|x|- modul de x la puterea modul de x

In punctul x=0 este:

a)0

b)+ infinit

c) -infinit

d) e

e)1

f)nu exista

Answer 1

$f(x) = \begin{cases} |x|^{|x|},\quad x\neq 0 \\ 1,\quad x = 0\end{cases}\\ \\\\ \underset{x<0}{\lim\limits_{x\to 0}}\,f'(x) =\underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{f(x)-f(0)}{x-0} = \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{(-x)^{-x}-1}{x}=\\ \\\\ y = (-x)^{-x}\\\\ \ln y = -x \ln(-x) \\\\ \dfrac{y'}{y} = -\ln(-x)-1 \\ \\ y' =-(-x)^{-x}\Big(\ln (-x) +1\Big)\\ \\ =\underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{-(-x)^{-x}\Big(\ln (-x) +1\Big)}{1} = -1\cdot (-\infty) = +\infty$

$\underset{x<0}{\lim\limits_{x\to 0}}\,f'(x) =\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{x^x-1}{x} \overset{(*)}{=}\\ \\ y = x^x \\\\ \ln y = x\ln x\\\\ \dfrac{y'}{y} = \ln x+1\\ \\ y' = y(\ln x+1) \\ \\y' = x^x(\ln x+1)\\ \\ \overset{(*)}{=} \underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{x^x(\ln x+1)-1}{1} =\dfrac{1\cdot(-\infty +1)-1}{1} = -\infty \\ \\ \\\Rightarrow f's \neq f'd \Rightarrow f'(0) \text{ nu exista}$

⇒ f) corect