Question

Qual é a derivada de raiz de 2x+1 Aplicando a definição

Answer 1

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


Aplicando a definição, a derivada de $f$ em cada ponto $x$ é dado por

$f'(x)=\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{f(x+h)-f(x)}{h}\\\\\\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{2(x+h)+1}-\sqrt{2x+1}}{h}\\\\\\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}$

( multiplicando e dividindo pelo conjugado )

$=\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}\cdot \dfrac{\sqrt{2x+2h+1}+\sqrt{2x+1}}{\sqrt{2x+2h+1}+\sqrt{2x+1}}\\\\\\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\big(\sqrt{2x+2h+1}-\sqrt{2x+1}\big)\cdot \big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}{h\cdot \big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}\\\\\\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{\big(\sqrt{2x+2h+1}\big)^2-\big(\sqrt{2x+1}\big)^2}{h\cdot \big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}\\\\\\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{2x+2h+1-(2x+1)}{h\cdot \big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}\\\\\\ =\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{2\diagup\!\!\!\! h}{\diagup\!\!\!\! h\cdot \big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}$

$=\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}\\\\\\ =\dfrac{2}{\sqrt{2x+2\cdot 0+1}+\sqrt{2x+1}}\\\\\\ =\dfrac{2}{\sqrt{2x+1}+\sqrt{2x+1}}\\\\\\ =\dfrac{\diagup\!\!\!\! 2}{\diagup\!\!\!\! 2\,\sqrt{2x+1}}\\\\\\ \therefore~~\boxed{\begin{array}{c} f'(x)=\dfrac{1}{\sqrt{2x+1}} \end{array}}~~~~\text{para }x\ne -\,\dfrac{1}{2}$


Bons estudos! :-)