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Derivada de $$$3 e^{- 4 r} \sin{\left(3 \theta \right)}$$$ con respecto a $$$r$$$
Calculadoras relacionadas: Calculadora de diferenciación logarítmica, Calculadora de diferenciación implícita con pasos
Tu aportación
Encuentra $$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)$$$.
Solución
Aplique la regla del múltiplo constante $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$ con $$$c = 3 \sin{\left(3 \theta \right)}$$$ y $$$f{\left(r \right)} = e^{- 4 r}$$$:
$${\color{red}\left(\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)\right)} = {\color{red}\left(3 \sin{\left(3 \theta \right)} \frac{d}{dr} \left(e^{- 4 r}\right)\right)}$$La función $$$e^{- 4 r}$$$ es la composición $$$f{\left(g{\left(r \right)} \right)}$$$ de dos funciones $$$f{\left(u \right)} = e^{u}$$$ y $$$g{\left(r \right)} = - 4 r$$$.
Aplicar la regla de la cadena $$$\frac{d}{dr} \left(f{\left(g{\left(r \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dr} \left(g{\left(r \right)}\right)$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(e^{- 4 r}\right)\right)} = 3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dr} \left(- 4 r\right)\right)}$$La derivada de la exponencial es $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dr} \left(- 4 r\right) = 3 \sin{\left(3 \theta \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dr} \left(- 4 r\right)$$Vuelva a la variable anterior:
$$3 e^{{\color{red}\left(u\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right) = 3 e^{{\color{red}\left(- 4 r\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right)$$Aplique la regla del múltiplo constante $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$ con $$$c = -4$$$ y $$$f{\left(r \right)} = r$$$:
$$3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(- 4 r\right)\right)} = 3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(- 4 \frac{d}{dr} \left(r\right)\right)}$$Aplique la regla de potencia $$$\frac{d}{dr} \left(r^{n}\right) = n r^{n - 1}$$$ con $$$n = 1$$$, en otras palabras, $$$\frac{d}{dr} \left(r\right) = 1$$$:
$$- 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(r\right)\right)} = - 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(1\right)}$$Por lo tanto, $$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$.
Respuesta
$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$A