Question

f(x) = x^2 - 2 lnx
f'(x) = ?

Rezultatul trebuie sa dea 2(x-1)(x+1) supra x

Answer 1

$f(x)=x^{2} -2lnx\\f'(x)=(x^{2} -2lnx)'=(x^{2})'-2(lnx)'=2x-2\frac{1}{x} =2x-\frac{2}{x} =\frac{2x^{2}-2 }{x} =\\=\frac{2(x^{2}-1) }{x} =\frac{2(x-1)(x+1)}{x}$

Bafta!

Answer 2

Identități folosite:

$(f \pm g)' = f'\pm g'$

$(x^n)' = nx^{n-1},\quad n-\text{ constant\u{a}}$

$\left[a\cdot u(x)\right]' = a\cdot u'(x),\quad a - \text{constant\u{a}}$

$(\ln x)' = \frac{1}{x}$

Rezolvare:

$f(x) = x^2-2\ln x$

$f'(x) = (x^2-2\ln x)' = (x^2)' - (2\ln x)'=$

$=(x^2)' - 2\cdot (\ln x)' = 2x^{2-1}-2\cdot \frac{1}{x} =$

$=2x-\frac{2}{x} =2(x-\frac{1}{x}) = \frac{2(x^2-1)}{x}=$

$=\frac{2(x-1)(x+1)}{x}$