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Derivada de $$$5 x^{x}$$$
Calculadoras relacionadas: Calculadora de diferenciación logarítmica, Calculadora de derivación implícita con pasos
Tu entrada
Halla $$$\frac{d}{dx} \left(5 x^{x}\right)$$$.
Solución
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 $$$c = 5$$$ y $$$f{\left(x \right)} = x^{x}$$$:
$${\color{red}\left(\frac{d}{dx} \left(5 x^{x}\right)\right)} = {\color{red}\left(5 \frac{d}{dx} \left(x^{x}\right)\right)}$$Utiliza la fórmula $$$f^{g{\left(x \right)}}{\left(x \right)} = e^{g{\left(x \right)} \ln\left(f{\left(x \right)}\right)}$$$ con $$$f{\left(x \right)} = x$$$ y $$$g{\left(x \right)} = x$$$ para reescribir la expresión compleja:
$$5 {\color{red}\left(\frac{d}{dx} \left(x^{x}\right)\right)} = 5 {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)}$$La función $$$e^{x \ln\left(x\right)}$$$ es la composición $$$f{\left(g{\left(x \right)} \right)}$$$ de dos funciones $$$f{\left(u \right)} = e^{u}$$$ y $$$g{\left(x \right)} = x \ln\left(x\right)$$$.
Aplica la regla de la cadena $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$5 {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)} = 5 {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(x \ln\left(x\right)\right)\right)}$$La derivada de la función exponencial es $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$5 {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$Volver a la variable original:
$$5 e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 e^{{\color{red}\left(x \ln\left(x\right)\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 x^{x} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$Aplica la regla del producto $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ con $$$f{\left(x \right)} = x$$$ y $$$g{\left(x \right)} = \ln\left(x\right)$$$:
$$5 x^{x} {\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = 5 x^{x} {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$La derivada del logaritmo natural es $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$5 x^{x} \left(x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right) = 5 x^{x} \left(x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right)$$Aplica la regla de la potencia $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ con $$$n = 1$$$, en otras palabras, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$5 x^{x} \left(\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1\right) = 5 x^{x} \left(\ln\left(x\right) {\color{red}\left(1\right)} + 1\right)$$Por lo tanto, $$$\frac{d}{dx} \left(5 x^{x}\right) = 5 x^{x} \left(\ln\left(x\right) + 1\right)$$$.
Respuesta
$$$\frac{d}{dx} \left(5 x^{x}\right) = 5 x^{x} \left(\ln\left(x\right) + 1\right)$$$A