Question

derivada pela definição de $\sqrt{x} / \sqrt{2$

Answer 1

$f(x) = \frac{ \sqrt{x} }{ \sqrt{2} }$


Racionalizando:

$f(x) = \frac{ \sqrt{x} }{ \sqrt{2} } \ . \ \frac{\sqrt{2}}{\sqrt{2}}$

$f(x) = \frac{ \sqrt{x} \ . \ \sqrt{2}}{ \sqrt{2} \ . \ \sqrt{2}}$

$f(x) = \frac{ \sqrt{2x} }{ 2 }$


Derivada pela definição de limite:

$f'(x) = \lim_{h \to \ 0} \ \frac{f(x + h)-f(x)}{h}$


$f'(x) = \lim_{h \to \ 0} \ \frac{\frac{ \sqrt{2(x + h)} }{ 2 } \ - \ \frac{ \sqrt{2x} }{ 2 }}{h}$

$f'(x) = \lim_{h \to \ 0} \ \frac{\frac{ \sqrt{2(x + h)} \ - \ \sqrt{2x}}{ 2 }}{h}$

$f'(x) = \lim_{h \to \ 0} \ \frac{ \sqrt{2(x + h)} \ - \ \sqrt{2x}}{ 2 } \ . \ \frac{1}{h}$

$f'(x) = \lim_{h \to \ 0} \ \frac{ \sqrt{2(x + h)} \ - \ \sqrt{2x}}{ 2h }$


Multiplicando pelo conjugado:

$f'(x) = \lim_{h \to \ 0} \ \frac{ \sqrt{2(x + h)} \ - \ \sqrt{2x}}{ 2h } \ . \ \frac{ \sqrt{2(x + h)} \ + \ \sqrt{2x}}{ \sqrt{2(x + h)} \ + \ \sqrt{2x} }$

$f'(x) = \lim_{h \to \ 0} \ \frac{(\sqrt{2(x + h)} \ - \ \sqrt{2x}) \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x})}{ 2h \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x}) }$

$f'(x) = \lim_{h \to \ 0} \ \frac{2(x + h) \ - \ 2x}{ 2h \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x}) }$

$f'(x) = \lim_{h \to \ 0} \ \frac{2x \ + \ 2h \ - \ 2x}{ 2h \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x}) }$

$f'(x) = \lim_{h \to \ 0} \ \frac{\not{2x} \ + \ 2h \ - \ \not{2x}}{ 2h \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x}) }$

$f'(x) = \lim_{h \to \ 0} \ \frac{2h}{ 2h \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x}) }$

$f'(x) = \lim_{h \to \ 0} \ \frac{\not{2h}}{ \not{2h} \ . \ (\sqrt{2(x + h)} \ + \ \sqrt{2x}) }$

$f'(x) = \lim_{h \to \ 0} \ \frac{1}{ \sqrt{2(x + h)} \ + \ \sqrt{2x} }$


Aplicando o limite:

$f'(x) = \ \frac{1}{ \sqrt{2(x \ + \ 0)} \ + \ \sqrt{2x} }$

$f'(x) = \ \frac{1}{ \sqrt{2x} \ + \ \sqrt{2x} }$

$f'(x) = \ \frac{1}{ 2\sqrt{2x} }$


Racionalizando:

$f'(x) = \ \frac{1}{ 2\sqrt{2x} } \ . \ \frac{ \sqrt{2x} }{\sqrt{2x} }$

$f'(x) = \ \frac{ \sqrt{2x} }{ 2\sqrt{2x} \ . \ \sqrt{2x} }$

$f'(x) = \ \frac{ \sqrt{2x} }{ 2 \ . \ 2x }$

$f'(x) = \ \frac{ \sqrt{2x} }{ 4x }$