Question

Seja f(x) = √x. Calcule f'(2).

Obs: aplicando a definição de derivada

Answer 1

Boa tarde Jailson!


Solução!


$f'(x)= \displaystyle \lim_{t \to 0} \dfrac{ (x_{0}-t)-f( x_{0})}{t}\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \dfrac{\sqrt {(x_{0}-t)} - \sqrt{( x_{0})}}{t} \times \frac{\sqrt{(x_{0}-t})+\sqrt{( x_{0})}}{\sqrt{(x_{0}-t)}+\sqrt{( x_{0})}}\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \frac{ (\sqrt{(x_{0}-t)})^{2} - (\sqrt{x_{0}}) ^{2} }{t.( \sqrt{x_{0}+t}). (\sqrt{x_{0} }})$



$f'(x)= \displaystyle \lim_{t \to 0} \frac{(x_{0}+t) - (x_{0}) }{t( \sqrt{x_{0}+t}. (\sqrt{x_{0} }}\\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \frac{(x_{0}) - (x_{0})+t }{t( \sqrt{x_{0}+t}. (\sqrt{x_{0} }}\\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \frac{t}{t( t\sqrt{x_{0}+t}. (\sqrt{x_{0} }}\\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \frac{1}{( \sqrt{x_{0}+t}). (\sqrt{x_{0}) }}$




$f'(x)= \displaystyle \lim_{t \to 0} \frac{1}{( \sqrt{x_{0}+0}). (\sqrt{x_{0}) }}\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \frac{1}{2( \sqrt{x_{0}) }}\\\\\\\ f'(x)= \displaystyle \lim_{t \to 0} \frac{1}{2\sqrt{x }}}$


$f'(2)= \displaystyle \lim_{t \to 0} \frac{1}{2\sqrt{x }}$



$Substiruindo!\\\\\\ \boxed{Resposta\displaystyle \lim_{t \to 0} \frac{1}{2\sqrt{2 }}}$



Boa tarde!
Bons estudos!