Question

Derivada por definición

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

Answer 1

la derivada de f'(x) esta definida por:

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

reemplazando datos:

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

multiplicando por el conjugado 
$\lim_{h \to \ 0} \frac{ \sqrt{2x+2h +1}- \sqrt{2x+1} }{h}* \frac{( \sqrt{2x+2h +1} + \sqrt{2x+1} )}{( \sqrt{2x+2h +1}+ \sqrt{2x+1} )}$

luego tenemos:

$\lim_{h \to \ 0} \frac{2h}{h*( \sqrt{2x+2h + 1}+ \sqrt{2x+1} }$

eliminando h y reemplazando el limite tenemos:

$\frac{2}{ \sqrt{2x+2*0+1}+ \sqrt{2x+1} }$

finalmente nos quedaria asi:

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

Answer 2

$\displaystyle f(x)=\sqrt{2x+1} \\[2pt] \Rightarrow \ f(x+h)=\sqrt{2(x+h)+1}=\sqrt{2x+2h+1} \\[8pt] f^{\prime }(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\[8pt] f^{\prime }(x)=\lim_{h\to 0} \frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h} \\[8pt] \text{Como es un l\'imite del tipo 0/0 ,se multiplica en este caso por la}\\ \text{conjugada del numerador.} \\[8pt] f^{\prime }(x)=\lim_{h\to 0}\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}\cdot \frac{\sqrt{2x+2h+1}+\sqrt{2x+1}}{\sqrt{2x+2h+1}+\sqrt{2x+1}}$

$\displaystyle \text{En el numerador se aplica diferencia de cuadrados }\\ (a+b)(a-b)=a^2-b^2 \\[4pt] =\lim_{h \to 0}\,\frac{\sqrt{2x+2h+1}^{\,2}-\sqrt{2x+1}^{\,2}}{h\left( \sqrt{2x+2h+1}+\sqrt{2x+1}\,\right)} \\[8pt] =\lim_{h \to 0}\,\frac{(2x+2h+1)-(2x+1)}{h\left( \sqrt{2x+2h+1}+\sqrt{2x+1}\,\right)} \\[8pt] =\lim_{h \to 0}\,\frac{2h}{h\left( \sqrt{2x+2h+1}+\sqrt{2x+1}\,\right)} \\[6pt] \text{Se simplifica el factor }h$

$\displaystyle =\lim_{h\to 0}\, \frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}} \\[8pt] =\frac{2}{\sqrt{2x+(2)(0)+1}+\sqrt{2x+1}}=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}\\[8pt] \text{Por lo tanto :} \\[8pt] f^{\prime }(x)=\frac{1}{\sqrt{2x+1}}$