derivada de 1/(x-4)sei que a resposta é 1/(x-4)² mas como se chega a isso ?.... Pergunta de ideia dejpfurlan

Eriivan

Verified answer

Sem usar a regra do quociente.

$f(x)= \frac{1}{(x-4)}\\
\\f(x)=(x-4)^{-1} \\
\\f'(x)=-1(x-4)^{-2}\\
\\\boxed{f'(x)=- \frac{1}{(x-4)^2} }$

3 votes Thanks 2

Eriivan fixed.

Niiya

Verified answer

$\boxed{\boxed{f'(x)=\lim\limits_{h\rightarrow 0}~\frac{f(x+h)-f(x)}{h}}}$
_____________________

$f(x)=1/(x-4)\\f(x+h)=1/(x+h-4)$
____

$f'(x)=\lim\limits_{h\rightarrow 0}~\frac{f(x+h)-f(x)}{h}}}$

$f'(x)=\lim\limits_{h\rightarrow 0}~\frac{\frac{1}{x+h-4}-\frac{1}{x-4}}{h}\\\\f'(x)=\lim\limits_{h\rightarrow 0}~\frac{\frac{x-4-(x+h-4)}{(x+h-4)(x-4)}}{h}\\\\f'(x)=\lim\limits_{h\rightarrow 0}~\frac{\frac{x-4-x-h+4}{(x+h-4)(x-4)}}{h}\\\\f'(x)=\lim\limits_{h\rightarrow 0}~\frac{\frac{-h}{(x+h-4)(x-4)}}{h}\\\\f'(x)=\lim\limits_{h\rightarrow 0}~\frac{-h}{h(x+h-4)(x-4)}\\\\f'(x)=\lim\limits_{h\rightarrow 0}~\frac{-1}{(x+h-4)(x-4)}\\\\f'(x)=\frac{-1}{(x+0-4)(x-4)}$

$f'(x)=-\frac{1}{(x-4)(x-4)}\\\\\boxed{\boxed{f'(x)=-\frac{1}{(x-4)^{2}}}}$

0 votes Thanks 1

Lista de comentários

Verified answer

Verified answer

Helpful Links

Helpful Social