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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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: \[ f'(x) = \frac{1}{\sqrt{2x + 1}} \]

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La derivada de \( f(x) = \sqrt{2x + 1} \) es \( f'(x) = \frac{1}{\sqrt{2x + 1}} \).
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