Question

Derivada da função 1/X^3, ultilizando a definição de derivada, ou seja, sem regras de derivação. Please, bitte, por favor, por favor, prego, obsecro, من فضلك, ju letum, per favor, qing, s'il vous plait.

Answer 1

$\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{f(x+h)-f(x)}{h}\\\\\\\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{\frac{1}{(x+h)^3}-\frac{1}{x^3}}{h}\\\\\\\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{\frac{x^3~-~(x+h)^3}{(x+h)^3\,.\,x^3}}{h}\\\\\\\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{x^3~-~(x+h)^3}{(x+h)^3\,.\,x^3~.~h}\\\\\\\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{x^3~-~(x^3+3x^2h+3xh^2+h^3)}{(x^3+3x^2h+3xh^2+h^3)\,.\,x^3~.~h}$

$\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{-3x^2h-3xh^2-h^3}{(x^3+3x^2h+3xh^2+h^3)\,.\,x^3~.~h}\\\\\\\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{h.(-3x^2-3xh-h^2)}{(x^3+3x^2h+3xh^2+h^3)\,.\,x^3~.~h}\\\\\\\frac{df(x)}{dx}~=~\underset{h\to0}{lim}~\frac{-3x^2-3xh-h^2}{(x^3+3x^2h+3xh^2+h^3)\,.\,x^3}\\\\\\\frac{df(x)}{dx}~=~\frac{-3x^2}{(x^3)\,.\,x^3}\\\\\\\frac{df(x)}{dx}~=~\frac{-3x^2}{x^6}\\\\\\\boxed{\frac{df(x)}{dx}~=~-\frac{3}{x^4}}$