f'(x) de f(x)=2 + x / 3 - x
Soluções para a tarefa
Respondido por
1
Regra do quociente:
![\boxed{\boxed{\dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^{2}}}} \boxed{\boxed{\dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^{2}}}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Cdfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%5Cdfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%5Cright%5D%3D%5Cdfrac%7Bf%27%28x%29g%28x%29-g%27%28x%29f%28x%29%7D%7B%5Bg%28x%29%5D%5E%7B2%7D%7D%7D%7D)
____________________________

____________________________
likarr:
não faz sentido
Perguntas interessantes
Português,
1 ano atrás
Biologia,
1 ano atrás
História,
1 ano atrás
História,
1 ano atrás
Matemática,
1 ano atrás
Matemática,
1 ano atrás