Question

Qual a derivada de raiz de x pela definição?

Answer 1

Boa Tarde!

$y= \sqrt{x}$
$y+$Δy=$\sqrt{x+\delta x}$

$\frac{\delta y}{\delta x} = \frac{ \sqrt{x+ \delta x}- \sqrt{x} }{\delta x}$

$lim \frac{\delta y}{\delta x} = lim \frac{ \sqrt{x+\delta x} - \sqrt{x} }{\delta x} * \frac{ \sqrt{x+\delta x}+ \sqrt{x} }{x+\delta x}+ \sqrt{x}}$
Δx→0

$\boxed { y'= \frac{1}{2 \sqrt{x} } }$

Bons estudos!!
=)

Answer 2

Olá

Derivada por definição

$\displaystyle \mathsf{ \lim_{h \to 0} ~ \frac{f(x+h)-f(x)}{h} }\\\\\\\mathsf{f(x)= \sqrt{x} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{ \sqrt{x+h} - \sqrt{x} }{h} }$

Temos que eliminar as raízes do numerador, para isso, basta multiplicar pelo conjugado.

$\displaystyle\mathsf{ \lim_{h \to 0} ~ \frac{ \sqrt{x+h} - \sqrt{x} }{h} ~\cdot~ \frac{ \sqrt{x+h}+ \sqrt{x} }{\sqrt{x+h}+ \sqrt{x} } }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{( \sqrt{x+h})^2 - (\sqrt{x})^2 }{h\cdot ( \sqrt{x+h}+ \sqrt{x} )}}\\\\\\\text{Cancela as raizes com o expoente}\\\\\mathsf{ \lim_{h \to 0} ~ \frac{ \diagup\!\!\!\!x+h- \diagup\!\!\!\!x }{h\cdot ( \sqrt{x+h}+ \sqrt{x} )} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{\diagup\!\!\!\! h }{\diagup\!\!\!h\cdot ( \sqrt{x+h}+ \sqrt{x} )} }$

$\displaystyle \mathsf{ \lim_{h \to 0} ~ \frac{1}{ \sqrt{x+h}+ \sqrt{x} } }~=~ \frac{1}{ \sqrt{x+0}+ \sqrt{x} } ~=~ \frac{1}{ \sqrt{x} + \sqrt{x} } ~=~\boxed{ \frac{1}{2 \sqrt{x} } }$