Derivada de $$$\ln\left(-1 + \frac{1}{x}\right)$$$
Calculadoras relacionadas: Calculadora de diferenciación logarítmica, Calculadora de derivación implícita con pasos
Tu entrada
Halla $$$\frac{d}{dx} \left(\ln\left(-1 + \frac{1}{x}\right)\right)$$$.
Solución
La función $$$\ln\left(-1 + \frac{1}{x}\right)$$$ es la composición $$$f{\left(g{\left(x \right)} \right)}$$$ de dos funciones $$$f{\left(u \right)} = \ln\left(u\right)$$$ y $$$g{\left(x \right)} = -1 + \frac{1}{x}$$$.
Aplica la regla de la cadena $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(-1 + \frac{1}{x}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(-1 + \frac{1}{x}\right)\right)}$$La derivada del logaritmo natural es $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(-1 + \frac{1}{x}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(-1 + \frac{1}{x}\right)$$Volver a la variable original:
$$\frac{\frac{d}{dx} \left(-1 + \frac{1}{x}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(-1 + \frac{1}{x}\right)}{{\color{red}\left(-1 + \frac{1}{x}\right)}}$$La derivada de una suma/diferencia es la suma/diferencia de las derivadas:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(-1 + \frac{1}{x}\right)\right)}}{-1 + \frac{1}{x}} = \frac{{\color{red}\left(- \frac{d}{dx} \left(1\right) + \frac{d}{dx} \left(\frac{1}{x}\right)\right)}}{-1 + \frac{1}{x}}$$Aplica la regla de la potencia $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ con $$$n = -1$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{1}{x}\right)\right)} - \frac{d}{dx} \left(1\right)}{-1 + \frac{1}{x}} = \frac{{\color{red}\left(- \frac{1}{x^{2}}\right)} - \frac{d}{dx} \left(1\right)}{-1 + \frac{1}{x}}$$La derivada de una constante es $$$0$$$:
$$\frac{- {\color{red}\left(\frac{d}{dx} \left(1\right)\right)} - \frac{1}{x^{2}}}{-1 + \frac{1}{x}} = \frac{- {\color{red}\left(0\right)} - \frac{1}{x^{2}}}{-1 + \frac{1}{x}}$$Simplificar:
$$- \frac{1}{x^{2} \left(-1 + \frac{1}{x}\right)} = \frac{1}{x \left(x - 1\right)}$$Por lo tanto, $$$\frac{d}{dx} \left(\ln\left(-1 + \frac{1}{x}\right)\right) = \frac{1}{x \left(x - 1\right)}$$$.
Respuesta
$$$\frac{d}{dx} \left(\ln\left(-1 + \frac{1}{x}\right)\right) = \frac{1}{x \left(x - 1\right)}$$$A