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