Derivada implícita de $$$x^{2} - x y + 3 y^{2} = 0$$$ con respecto a $$$x$$$
Tu aportación
Encuentra $$$\frac{d}{dx} \left(x^{2} - x y + 3 y^{2} = 0\right)$$$.
Solución
Derive por separado ambos lados de la ecuación (trate a $$$y$$$ como una función de $$$x$$$ ): $$$\frac{d}{dx} \left(x^{2} - x y{\left(x \right)} + 3 y^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(0\right)$$$.
Diferenciar el LHS de la ecuación.
La derivada de una suma/diferencia es la suma/diferencia de derivadas:
$${\color{red}\left(\frac{d}{dx} \left(x^{2} - x y{\left(x \right)} + 3 y^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) + \frac{d}{dx} \left(3 y^{2}{\left(x \right)}\right)\right)}$$Aplique la regla del múltiplo constante $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ con $$$c = 3$$$ y $$$f{\left(x \right)} = y^{2}{\left(x \right)}$$$:
$${\color{red}\left(\frac{d}{dx} \left(3 y^{2}{\left(x \right)}\right)\right)} + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) = {\color{red}\left(3 \frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$La función $$$y^{2}{\left(x \right)}$$$ es la composición $$$f{\left(g{\left(x \right)} \right)}$$$ de dos funciones $$$f{\left(u \right)} = u^{2}$$$ y $$$g{\left(x \right)} = y{\left(x \right)}$$$.
Aplicar la regla de la cadena $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$3 {\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 3 {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$Aplique la regla de potencia $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ con $$$n = 2$$$:
$$3 {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 3 {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$Vuelva a la variable anterior:
$$6 {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 6 {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$Aplique la regla de potencia $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ con $$$n = 2$$$:
$$6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right) = 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + {\color{red}\left(2 x\right)} - \frac{d}{dx} \left(x y{\left(x \right)}\right)$$Aplique 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 $$$g{\left(x \right)} = y{\left(x \right)}$$$:
$$2 x + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(\frac{d}{dx} \left(x y{\left(x \right)}\right)\right)} = 2 x + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(\frac{d}{dx} \left(x\right) y{\left(x \right)} + x \frac{d}{dx} \left(y{\left(x \right)}\right)\right)}$$Aplique la regla de potencia $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ con $$$n = 1$$$, en otras palabras, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x - y{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) = - x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x - y{\left(x \right)} {\color{red}\left(1\right)} + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right)$$Por lo tanto, $$$\frac{d}{dx} \left(x^{2} - x y{\left(x \right)} + 3 y^{2}{\left(x \right)}\right) = - x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x + 6 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - y{\left(x \right)}.$$$
Derive la RHS de la ecuación.
La derivada de una constante es $$$0$$$:
$${\color{red}\left(\frac{d}{dx} \left(0\right)\right)} = {\color{red}\left(0\right)}$$Por lo tanto, $$$\frac{d}{dx} \left(0\right) = 0$$$.
Por tanto, hemos obtenido la siguiente ecuación lineal con respecto a la derivada: $$$- x \frac{dy}{dx} + 2 x + 6 y \frac{dy}{dx} - y = 0$$$.
Resolviéndola, obtenemos que $$$\frac{dy}{dx} = \frac{2 x - y}{x - 6 y}$$$.
Respuesta
$$$\frac{dy}{dx} = \frac{2 x - y}{x - 6 y}$$$A