# Partial derivative and chain rule

1. Feb 9, 2013

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

File size:
5.2 KB
Views:
89
2. Feb 9, 2013

### 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.