Multivariable Chain rule for higher order derivatives

[Sunfire]

Hello,

Given is the function

f = f(a,b,t), where a=a(b) and b = b(t). Need to express first and second order derivatives.

\frac{\partial f}{\partial a} and \frac{\partial f}{\partial b} should be just that, nothing more to it here, correct?

But

\frac{df}{dt} = \frac{\partial f}{\partial a} \frac{da}{db} \frac{db}{dt} + \frac{\partial f}{\partial b} \frac{db}{dt} + \frac{\partial f}{\partial t}, by the chain rule, correct?

I need to express \frac{\partial f}{\partial t}, but the above chain rule puts the total derivative \frac{df}{dt} in the expression and it gets messy. I mean, how do I express

\frac{\partial f}{\partial t}?

Then I need also \frac{\partial^2 f}{\partial t^2}, \frac{\partial^2 f}{\partial a^2} and \frac{\partial^2 f}{\partial b^2}.

Anyone well versed in partial derivatives?

 

[Sunfire]

Let me try to clear it up a bit

Let f = f(a,b,t) where a=a(b(t)), b=b(t)

Then

\frac{df}{dt}=\frac{∂f}{∂a}\frac{da}{db}\frac{db}{dt} + \frac{∂f}{∂b}\frac{db}{dt} + \frac{∂f}{∂t}, correct?

How does one express

\frac{d^2f}{dt^2}=?

 

[Sunfire]

Okay, Let me simplify this to

Let f = f(a,b) where a=a(b)

Then

\frac{df}{db}=\frac{∂f}{∂a}\frac{da}{db} + \frac{∂f}{∂b}, correct?

How does one express

\frac{d^2f}{db^2}=?