# Chain Rule in Lagrangian Transformation

Hello,
I'm trying to follow Goldstein textbook to show that the Lagrangian is invariant under coordinate transformation. I got confused by the step below
So
## L = L(q_{i}(s_{j},\dot s_{j},t),\dot q_{i}(s_{j},\dot s_{j},t),t)##

The book shows that ##\dot q_{i} = \frac {\partial q_{i}}{\partial s_{j}} \dot s_{j} + \frac{\partial q_{i}}{\partial t} ##,and I'm not sure where the ##\frac{\partial q_{i}}{\partial t} ## term come from? I've tried to look up chain rules for coordinate transformation but I can't find anything.

Thank you :D

Orodruin
Staff Emeritus
Homework Helper
Gold Member
It is just the standard chain rule. Consider a function ##f(h,g)## where ##h## and ##g## are functions of some parameter ##t## and differentiate ##f## with respect to ##t##. The chain rule now states
$$\frac{df}{dt} = \frac{\partial f}{\partial h} \frac{dh}{dt} + \frac{\partial f}{\partial g} \frac{dg}{dt}.$$
Now, if ##g = t## you would obtain ##dg/dt = dt/dt = 1## and ##\partial f/\partial g = \partial f/\partial t## and therefore
$$\frac{df}{dt} = \frac{\partial f}{\partial h} \frac{dh}{dt} + \frac{\partial f}{\partial t}.$$

What is unclear is why a term ##(\partial q/\partial \dot s) \ddot s## is not included. You can only ignore this term if ##q## is a function only of ##s## and not of ##\dot s##.

SEGFAULT1119
@Oroduin,

You are correct. q is independent of ##\dot s##. I mistyped the question. And thank you for your answer.

vanhees71