questão aparentemente simples, mas tô aprendendo agora...
2/3 (5x+3 / 5x-3)
gabarito: -20 / (5x-3)^2
vamos la
f(x)= 2/3(5x+3)/(5x-3) para ficar mais facil chamaremos 2/3 de k
f(x)= k(5x+3)/(5x-3)
agora aplicaremos a regra do quociente
f'(x)= k[(5x+3)'(5x-3)-(5x-3)'(5x+3)]/(5x-3)²
f'(x)= k[(5(5x-3)-(5)(5x+3)]/(5x-3)²
f'(x)= k(25x-15-25x-15)/(5x-3)²
f'(x)= 2/3(-30/(5x-3)²
f'(x)= -20/(5x-3)² okkkkk
Usando a fórmula do quociente:
y' = (vu' - uv')/v^2
y' = 2/3{[(5x-3)5 - (5x+3)5]}/(5x-3)^2 = 2/3[(25x - 15) - (25x + 15)]/(5x - 3)^2 = 2/3(-30)/(5x-3)^2 = -20/(5x-3)^2.
Comments
vamos la
f(x)= 2/3(5x+3)/(5x-3) para ficar mais facil chamaremos 2/3 de k
f(x)= k(5x+3)/(5x-3)
agora aplicaremos a regra do quociente
f'(x)= k[(5x+3)'(5x-3)-(5x-3)'(5x+3)]/(5x-3)²
f'(x)= k[(5(5x-3)-(5)(5x+3)]/(5x-3)²
f'(x)= k(25x-15-25x-15)/(5x-3)²
f'(x)= 2/3(-30/(5x-3)²
f'(x)= -20/(5x-3)² okkkkk
Usando a fórmula do quociente:
y' = (vu' - uv')/v^2
y' = 2/3{[(5x-3)5 - (5x+3)5]}/(5x-3)^2 = 2/3[(25x - 15) - (25x + 15)]/(5x - 3)^2 = 2/3(-30)/(5x-3)^2 = -20/(5x-3)^2.