$$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} \sin \theta+ \frac{\partial f}{\partial v} \cos \theta$$
$$\frac{\partial^2f}{\partial y^2}=\frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial y} \right)=\frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right)= \frac{\partial}{\partial y} \frac{\partial f}{\partial u} \sin \theta + \frac{\partial}{\partial y} \frac{\partial f}{\partial v} \cos \theta$$
Ahora el problema es calcular $\frac{\partial}{\partial y} \frac{\partial f}{\partial u}$ y $\frac{\partial } {\partial y} \frac{\partial f}{\partial v}$ . Podemos pensar en $\frac{\partial}{\partial y} \frac{\partial f}{\partial u}$ como $\frac{\partial}{\partial u} \frac{\partial f}{\partial y}$ y de forma similar para este último.
$$\frac{\partial}{\partial u} \frac{\partial f}{\partial y} = \frac{\partial}{\partial u} \left( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right) = \frac{\partial^2 f}{\partial u^2} \sin \theta + \frac{\partial f}{\partial u \partial v} \cos \theta$$
$$\frac{\partial}{\partial v} \frac{\partial f}{\partial y} = \frac{\partial}{\partial v} \left( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right) = \frac{\partial^2 f}{\partial u \partial v} \sin \theta + \frac{\partial^2 f}{\partial v^2} \cos \theta$$
Ahora, conecte esto de nuevo a $\frac{\partial^2f}{\partial y^2}$ dando,
$$\frac{\partial^2f}{\partial y^2}= \left (\frac{\partial^2 f}{\partial u^2} \sin \theta + \frac{\partial f}{\partial u \partial v} \cos \theta \right) \sin \theta + \left (\frac{\partial^2 f}{\partial u \partial v} \sin \theta + \frac{\partial^2 f}{\partial v^2} \cos \theta \right) \cos \theta $$
Repita el proceso para $\frac{\partial^2 f}{\partial x^2}$ , sumadlo todo y llegaréis al resultado deseado.
1 votos
Tienes la idea correcta, pero en lugar de escribir $d/dx = d/du du/dx$ empezar con la derivada interior. Así, primero $df/dx = df/du du/dx$ y luego usar la regla de la cadena. Terminas con el paso que anotaste.
4 votos
Posible duplicado de Demostración de que la ecuación de Laplace es rotacionalmente invariante utilizando la regla de la cadena