Question

F(x)= x²+2x+1 totul supra x+2. Aflati f'(x)

Answer 1

$f(x) = \frac{x^2 + 2x + 1}{x+2} = \frac{(x+1)^2}{x+2}\\(\frac{f}{g})' = \frac{f'g - fg'}{g^2} \\f'(x) = \frac{(x+2)[(x+1)^2]' - (x+1)^2[x+2]'}{(x+2)^2} = \frac{(x+2)\cdot 2(x+1)\cdot [x+1]' - (x+1)^2\cdot 1}{(x+2)^2} = \frac{2(x+1)(x+2) - (x+1)^2}{(x+2)^2} = (x+1)\cdot \frac{2(x+2) - (x+1)}{(x+2)^2} = (x+1)(\frac{2(x+2)}{(x+2)^2} - \frac{x+1}{(x+2)^2}) = (x+1)(\frac{2}{x+2} - \frac{(x+1)}{(x+2)^2}) = \frac{x+1}{x+2}(2 - \frac{x+1}{x+2})$

Answer 2

Explicație pas cu pas: