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