Question

Tentei fazer a derivada de F(x)=$<var>\sqrt{x+1}</var>$, X>-1

ficou assim $<var>\lim_{x \to \.a}} \frac{f(x)-f(a)}{x-a}</var>$

$<var>\lim_{x \to \.a}} \frac{\sqrt{x+1}-\sqrt{a+1}}{x-a}</var>$

$<var>\lim_{x \to \.a}} \frac{\sqrt{x+1}-\sqrt{a+1}}{x-a}.\frac{\sqrt{x+1}+\sqrt{a+1}}{\sqrt{x+1}+\sqrt{a+1}}</var>$

$<var>\lim_{x \to \.a}}\frac{x+1-a+1}{(x-a).(\sqrt{x+1}+\sqrt{a+1})}=\frac{x-a+2}{(x-a).(\sqrt{x+1}+\sqrt{a+1})}</var>$

Travei aqui D: Alguem poderia dar um help?

Answer 1

Oi aureo,

 

vc errou só um sinal

 

$<var>\lim_{x \to \.a}} \frac{\sqrt{x+1}-\sqrt{a+1}}{x-a}.\frac{\sqrt{x+1}+\sqrt{a+1}}{\sqrt{x+1}+\sqrt{a+1}}\\ \\ </var>$

 

$<var>\lim_{x \to \.a}}\frac{x+1-a-1}{(x-a).(\sqrt{x+1}+\sqrt{a+1})}=\frac{x-a}{(x-a).(\sqrt{x+1}+\sqrt{a+1})}</var>$

 

cortará (x-a) com (x-a)

 

e ficará

 

$<var>\lim_{x \to \.a}}=\frac{1}{(\sqrt{x+1}+\sqrt{a+1})} </var>$

$<var>\frac{1}{\sqrt{x+1}+\sqrt{x+1}}\\ \\ \frac{1}{2\sqrt{x+1}}\\</var>$