Norton Frank
02/28/2024 · Senior High School
\( \Rightarrow \begin{array}{l}\sqrt{\mathrm{x}+1+\mathrm{x}} \\ \text { Derivar por definición }\end{array} \)
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Step-by-step Solution
Para derivar la función \( f(x) = \sqrt{x + 1 + x} \) por definición, primero simplificamos la función:
\[
f(x) = \sqrt{2x + 1}
\]
La derivada de una función \( f(x) \) por definición se calcula como:
\[
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
\]
Sustituyendo \( f(x) \):
\[
f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) + 1} - \sqrt{2x + 1}}{h}
\]
Simplificamos el numerador:
\[
f'(x) = \lim_{h \to 0} \frac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h}
\]
Para simplificar la expresión, multiplicamos el numerador y el denominador por el conjugado del numerador:
\[
f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{2x + 2h + 1} - \sqrt{2x + 1}\right)\left(\sqrt{2x + 2h + 1} + \sqrt{2x + 1}\right)}{h\left(\sqrt{2x + 2h + 1} + \sqrt{2x + 1}\right)}
\]
Esto nos da:
\[
f'(x) = \lim_{h \to 0} \frac{(2x + 2h + 1) - (2x + 1)}{h\left(\sqrt{2x + 2h + 1} + \sqrt{2x + 1}\right)}
\]
Simplificando el numerador:
\[
f'(x) = \lim_{h \to 0} \frac{2h}{h\left(\sqrt{2x + 2h + 1} + \sqrt{2x + 1}\right)}
\]
Cancelamos \( h \) en el numerador y el denominador:
\[
f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}
\]
Ahora, evaluamos el límite cuando \( h \) tiende a 0:
\[
f'(x) = \frac{2}{\sqrt{2x + 1} + \sqrt{2x + 1}} = \frac{2}{2\sqrt{2x + 1}} = \frac{1}{\sqrt{2x + 1}}
\]
Por lo tanto, la derivada de \( f(x) = \sqrt{2x + 1} \) es:
\[
f'(x) = \frac{1}{\sqrt{2x + 1}}
\]
Quick Answer
La derivada de \( f(x) = \sqrt{2x + 1} \) es \( f'(x) = \frac{1}{\sqrt{2x + 1}} \).
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