ProfdeMaths1

Verified answer

Il y a bcp de fonctions...

1) f(x)=x-√(x²+2) ; Df=IR
f'(x)=1-2x/(2√(x²+2))=1-x/√(x²+2)

2) f(x)=2x²-√x ; Df=IR+
f'(x)=4x-1/(2√x)

3) f(x)=(x^4-x²)/(x²+1) ; Df=IR
f'(x)=((4x³-2x)(x²+1)-2x(x^4-x²))/(x²+1)²
      =(4x^5+4x³-2x³-2x-2x^5+2x³)/(x²+1)²
      =(2x^5+4x³-2x)/(x²+1)²

4) f(x)=x³/(2x-1)² ; Df=IR\{1/2}
f'(x)=(3x²(2x-1)²-x³(8x-4))/(2x-1)^4
      =(3x²(2x-1)-4x³)/(2x-1)³
      =(2x³-3x²)/(2x-1)³

taalbabachir 1) f(x) = x - √x² + 2    Df = R

Lim f(x) = - ∞  Lim f(x) = - ∞
x→ - ∞            x→ +∞

f '(x) = 1 -  1/2 (2 x)/√x² + 2 = 1 -  x/√x² + 1

2) f(x) = 2 x² - √x     Df = [0 ; + ∞[

 Lim f(x) = 0    Lim f(x) = + ∞
 x→0               x→ +∞ 

 f '(x) = 4 x - 1/2(√x)

 3) f(x) = x⁴ - x²)/x² + 1   Df = R


f '(x) = (4 x³ - 2 x)(x² + 1) - 2 x(x⁴ - x²)]/(x² + 1)²
 
 f '(x) = 4 x⁵ + 4 x³ - 2 x³ - 2 x - 2 x(x⁶ + x⁴ - x⁴ - x²)

        = 4 x⁵ + 4 x³ - 2 x - 2 x⁷

 f '(x) = - 2 x⁷ + 4 x⁵ + 4 x³ - 2 x)/(x² + 1)²

 vous continuez