Partial derivative and chain rule

[rsaad]

How is the double derivative equal to that in the equation 2 in the attachment? =|

 

[Simon Bridge]

$$\dot{y} = \frac{\partial f_i}{\partial x_j}\dot{x_j} + \frac{\partial f_i}{\partial t}\qquad \text{...(1)}\\

\ddot{y} = \frac{\partial f_i}{\partial x_j}\ddot{x_j} + \frac{\partial^2 f_i}{\partial x_j \partial x_k}\dot{x_j}\dot{x_k} + 2\frac{\partial^2 f_i}{\partial x_j \partial t}\dot{x_j} + \frac{\partial^2 f_i}{\partial t^2}\qquad \text{...(2)}$$​

2 follows from 1 (and the definition of y - what is this?) by the chain rule ... so apply the chain rule and show where you get stuck.