Question

Calcular a derivada de:
$z= \frac{2x+1}{ \sqrt{x} }$

Answer 1

$z= \frac{2x+1}{\sqrt{x}} \\\\z= \frac{2x}{ \sqrt{x} } + \frac{1}{ \sqrt{x} } \\\\z= \frac{2x}{x^{ \frac{1}{2} }}+ \frac{1}{x^{ \frac{1}{2} }} \\\\z=2x^{(1- \frac{1}{2}) } + x^{ \frac{-1}{2} }\\\\\boxed{ z=2x^{ \frac{1}{2}} +x^{ \frac{-1}{2} }}}\\\\\text{derivando}\\\\ z'=2* \frac{1}{2}*x^{ (\frac{1}{2}-1) }+ \frac{-1}{2}*x^{ (\frac{-1}{2}-1) } \\\\z'= x^{ \frac{-1}{2} }- \frac{1}{2}x^{ \frac{-3}{2} }$

$z'= \frac{1}{x^{ \frac{1}{2} }} - \frac{1}{x^{ \frac{3}{2} }} \\\\ \boxed{\boxed{z'= \frac{1}{ \sqrt{x} } - \frac{1}{2 \sqrt{x^3} } }}$

Answer 2

Olá

Temos que usar a regra do quociente, dada por $z'= \frac{f'\cdot g-f\cdot g'}{ g^2 }$

A derivada de 2x+1 é 2
A derivada de √x é $\frac{1}{2 \sqrt{x} }$


$z= \frac{2x+1}{ \sqrt{x} } \\ \\ z'= \frac{2( \sqrt{x} )~-~(2x+1)\cdot \frac{1}{2 \sqrt{x} } }{ (\sqrt{x})^2 } \\ \\ Cancela~a~raiz~com~o~expoente~no~denominador \\ \\ z'= \frac{2( \sqrt{x} )~-~(2x+1)\cdot \frac{1}{2 \sqrt{x} } }{ x } \\ \\ z'= \frac{2( \sqrt{x} )~-~ \frac{2x+1}{2 \sqrt{x} } }{ x } \\ \\ Faz~o~mmc~e~agrupa~os~2~termos \\ \\ z'= \frac{ \frac{(2 \sqrt{x})^2~-~(2x-1) }{2 \sqrt{x} } }{x} \\ \\ z'= \frac{ \frac{4x~-~2x-1 }{2 \sqrt{x} } }{x}$


$z'= \frac{ \frac{2x-1 }{2 \sqrt{x} } }{x} \\ \\ Divisao~de~fracoes~,~multiplica~pelo~inverso \\ \\ z'= \frac{ 2x-1 }{2 \sqrt{x} } \cdot \frac{1}{x} \\ \\ Transforma~a~ \sqrt{x} ~em~fracao~para~simplificarmos \\ \\ z'= \frac{ 2x-1 }{2 (x)^ \frac{1}{2} } \cdot \frac{1}{x} \\ \\ As~bases~sao~iguais,~soma~os~expoentes \\ \\ z'= \frac{ 2x-1 }{2 (x)^ \frac{1}{2}^+^1 } \\ \\ z'= \frac{ 2x-1 }{2 (x)^ \frac{3}{2} } \\ \\ Volta~pra~forma~de~raiz$

$\boxed{\boxed{z'=\frac{ 2x-1 }{2 \sqrt{x^3} }}}$