SENDO F(X)=2X^2/3X-5 A DERIVADA F"(2) É IGUAL A:

Resposta:

f(x) = 2x²/(3x - 5)

Primeira derivada:

f'(x) = [2x²/(3x - 5)]'

Aplique a regra do quociente: (f/g)' = (f'.g - g'.f)/g²

f'(x) = [(2x²)'.(3x - 5) - (3x - 5)'.(2x²)]/(3x - 5)²

f'(x) = [(2x²)'.(3x - 5) - ((3x)' - (5)').(2x²)]/(3x - 5)²

f'(x) = [4x.(3x - 5) - (3 - 0).(2x²)]/(3x - 5)²

f'(x) = [4x.(3x - 5) - 3.2x²]/(3x - 5)²

f'(x) = (12x² - 20x - 6x²)/(3x - 5)²

f'(x) = (6x² - 20x)/(3x - 5)²

Segunda derivada:

Aplique novamente a regra do quociente:

f''(x) = [(6x² - 20x)'.(3x - 5)² - ((3x - 5)²)'.(6x² - 20x)]/[(3x - 5)²]²

f''(x) = [((6x²)' - (20x)').(3x - 5)² - ((3x - 5)²)'.(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - ((3x - 5)²)'.(6x² - 20x)]/(3x - 5)⁴

Faça u = 3x - 5 e aplique a regra da cadeia.

f''(x) = [(12x - 20).(3x - 5)² - (u²)'.(u)'.(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - 2u.(u)'.(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - 2.(3x - 5).(3x - 5)'.(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - 2.(3x - 5).((3x)' - (5)').(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - 2.(3x - 5).(3 - 0).(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - 2.(3x - 5).3.(6x² - 20x)]/(3x - 5)⁴

f''(x) = [(12x - 20).(3x - 5)² - 6.(3x - 5).(6x² - 20x)]/(3x - 5)⁴

Para x = 2:

f''(2) = [(12.2 - 20).(3.2 - 5)² - 6.(3.2 - 5).(6.2² - 20.2)]/(3.2 - 5)⁴

f''(2) = [(24 - 20).(6 - 5)² - 6.(6 - 5).(6.4 - 40)]/(6 - 5)⁴

f''(2) = [4.1² - 6.1.(24 - 40)]/1⁴

f''(2) = (4.1 + 6.16)/1

f''(2) = 4 + 96

f''(2) = 100