Question

x^2 -1/ x^2 totul derivat​

Answer 1

$\Big(\dfrac{x^2-1}{x^2}\Big)' = \Big(\dfrac{x^2}{x^2} - \dfrac{1}{x^2}\Big)' = \Big(1-\dfrac{1}{x^2}\Big)' =\\ \\ = 0-\dfrac{0\cdot x^2 -1\cdot 2x}{(x^2)^2} = \dfrac{2x}{x^4} = \dfrac{2}{x^3}$

Answer 2

Răspuns:

$\frac{2}{x^3}$

Explicație pas cu pas:

Metoda 1:

Folosim formula: $(\frac{f(x)}{g(x)})'=\frac{f'(x)*g(x)-f(x)*g'(x)}{g^2(x)}$.

$(\frac{x^2-1}{x^2})'=\frac{(x^2-1)'*x^2-(x^2-1)*(x^2)'}{(x^2)^2}=\frac{2x*x^2-2x*(x^2-1)}{x^4}=\frac{2x^3-2x^3+2x}{x^4}=\frac{2x}{x^4}=\frac{2}{x^3}$

Metoda 2:

Distribuim numaratorul pentru a forma doua fractii.

Folosim formulele: $[f(x)-g(x)]'=f'(x)-g'(x)$, $1'=0$ si $(\frac{1}{f(x)})=-\frac{1}{f^2(x)}*f'(x)$.

$(\frac{x^2-1}{x^2})'=(\frac{x^2}{x^2}-\frac{1}{x^2})'=(1-\frac{1}{x^2})'=1'-(\frac{1}{x^2})'=-(-\frac{1}{(x^2)^2})*(x^2)'=\frac{2x}{x^4}=\frac{2}{x^3}$

Metoda 3:

Distribuim numaratorul pentru a forma doua fractii.

Folosim formulele: $[f(x)-g(x)]'=f'(x)-g'(x)$, $1'=0$, $\frac{1}{x^n}=x^{-n}$ si $[f^n(x)]'=n*f^{n-1}(x)*f'(x)$.

$(\frac{x^2-1}{x^2})'=(\frac{x^2}{x^2}-\frac{1}{x^2})'=(1-\frac{1}{x^2})'=1'-(\frac{1}{x^2})'=-(x^{-2})'=-(-2)*x^{-2-1}=2x^{-3}=\frac{2}{x^3}$

Metoda 4:

Distribuim numaratorul pentru a forma doua fractii.

Folosim formulele: $[f(x)-g(x)]'=f'(x)-g'(x)$, $1'=0$ si $(\frac{f(x)}{g(x)})'=\frac{f'(x)*g(x)-f(x)*g'(x)}{g^2(x)}$.

$(\frac{x^2-1}{x^2})'=(\frac{x^2}{x^2}-\frac{1}{x^2})'=(1-\frac{1}{x^2})'=1'-(\frac{1}{x^2})'=(-1)*\frac{1'*x^2-1*(x^2)'}{(x^2)^2}=(-1)*\frac{-2x}{x^4}=\frac{2x}{x^4}=\frac{2}{x^3}$