Question

Poate sa imi arate cineva cum se calculeaza aceasta derivata?

Answer 1

Am atasat rezolvarea.

8b833b8737462f4be43ddbb5187f21cd.pdf

Answer 2

$\displaystyle \mathtt{f:(0,\infty)\rightarrow \mathbb{R},~f(x)=ln~x- \frac{2(x-1)}{x+1} }\\ \\ \mathtt{f'(x)=\left(ln~x- \frac{2(x-1)}{x+1}\right)'=(ln~x)'-\left( \frac{2(x-1)}{x+1}\right)'= }\\ \\ \mathtt{=(ln~x)'-2\left( \frac{x-1}{x+1}\right)'= \frac{1}{x}-2\cdot \frac{(x-1)' \cdot (x+1)-(x-1)\cdot(x+1)'}{(x+1)^2}= }\\ \\ \mathtt{= \frac{1}{x}-2 \cdot \frac{(x'-1') \cdot (x+1)-(x-1) \cdot (x'+1')}{(x+1)^2}=}$
$\displaystyle \mathtt{= \frac{1}{x}-2 \cdot \frac{(1-0) \cdot(x+1)-(x-1)\cdot(1+0)}{(x+1)^2}= }\\ \\ \mathtt{= \frac{1}{x}-2 \cdot \frac{1 \cdot (x+1)-(x-1) \cdot 1}{(x+1)^2} = \frac{1}{x}-2 \cdot \frac{x+1-(x-1)}{(x+1)^2}= }\\ \\ \mathtt{= \frac{1}{x}-2 \cdot \frac{x+1-x+1}{(x+1)^2}= \frac{1}{x}-2 \cdot \frac{2}{(x+1)^2} = \frac{1}{x}- \frac{4}{(x+1)^2}=}\\ \\ \mathtt{= \frac{(x+1)^2-4x}{x(x+1)^2}= \frac{x^2+2 \cdot x \cdot 1+1^2-4x}{x(x+1)^2}= \frac{x^2+2x+1-4x}{x(x+1)^2} =}$
$\displaystyle \mathtt{= \frac{x^2-2x+1}{x(x+1)^2}= \frac{(x-1)^2}{x(x+1)^2} }$