Question

Calcule a derivada da função abaixo pela definição de derivada (pelo limite).
f(x) = √x

Answer 1

A função em questão é:

$f\left(x\right)=\sqrt{x}$

Aplicando a derivada da função pelo limite:

$\lim _{h\to 0}\left(\dfrac{f\left(x+h\right)-f\left(x\right)}{h}\right)\\\\\lim _{h\to 0}\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\right)\\\\\lim _{h\to 0}\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\\\\\lim _{h\to 0}\left(\dfrac{\left(\sqrt{x+h}\right)^2-\left(\sqrt{x}\right)^2}{h\cdot \left(\sqrt{x+h}+\sqrt{x}\right)}\right)\\\\\lim _{h\to 0}\left(\dfrac{x+h-x}{h\cdot \left(\sqrt{x+h}+\sqrt{x}\right)}\right)$
$\lim _{h\to 0}\left(\dfrac{h}{h\cdot \left(\sqrt{x+h}+\sqrt{x}\right)}\right)\\\\\lim _{h\to 0}\left(\dfrac{1}{\sqrt{x+h}+\sqrt{x}}\right)$

Substituindo h = 0:

$\dfrac{1}{\sqrt{x+0}+\sqrt{x}}\\\\\dfrac{1} {\sqrt{x}+\sqrt{x}}\\\\\boxed{\bold{\frac{1}{2\sqrt{x}}}}$
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Resposta: $\bold{f'\left(x\right)=\dfrac{1}{2\sqrt{x}}}$