Multivariable Chain rule for higher order derivatives

Sunfire
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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.

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

But

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

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

[itex]\frac{\partial f}{\partial t}[/itex]?

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

Anyone well versed in partial derivatives?
 
on Phys.org
Let me try to clear it up a bit

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

Then

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

How does one express

[itex]\frac{d^2f}{dt^2}[/itex]=?
 
Okay, Let me simplify this to

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

Then

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

How does one express

[itex]\frac{d^2f}{db^2}[/itex]=?
 

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