Second order partial derivative

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SUMMARY

The discussion focuses on calculating the second order partial derivative with respect to the variable z, where a and b are functions of z. The correct application of the chain rule is emphasized, specifically the expression w' = faa' + fbb', where w = f(a, b). The participants clarify that the second order derivative should be expressed in terms of the derivatives of a and b, leading to the formulation w'' = (fa)'a' + faa'' + (fb)'b' + fbb''. The use of proper notation and clarity in defining functions is highlighted as essential for accurate differentiation.

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intervoxel
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Homework Statement


a and b are functions of z:

a=a(z); b=b(z)

I want to calculate the second order partial derivative operator on z

Homework Equations


Using the chain rule:

[tex] \frac{\partial}{\partial z}=\frac{\partial a}{\partial z}\frac{\partial}{\partial a}+\frac{\partial b}{\partial z}\frac{\partial}{\partial b}[/tex]

The Attempt at a Solution


Is it correct?
[tex] \frac{\partial^2}{\partial z^2}=\frac{\partial}{\partial a^2}+2\frac{\partial^2}{\partial a\partial b}+\frac{\partial^2}{\partial b^2}[/tex]
 
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Your problem is incompletely stated and uses poor notation and I think the answer is no anyway. It would be much clearer to use subscript notation for partial derivatives and ' for ordinary derivatives. Here is what I am guessing you are asking.

Let w = f(a,b) where a and b are functions of z. Differentiating with respect to z, the derivatives of a and b would be a' and b', not partial derivatives. The derivatives of w with respect to a and b would be partials: wa and wb. The derivative of w with respect to z would be w'. The chain rule gives:

w' = faa' + fbb'

This is your chain rule you have given in the relevant equations. Now you want to calculate w''.

w'' = (faa' + fbb')'

Can you take it from there?

= (fa)'a' + faa'' + (fb)'b' + fbb''
 
intervoxel said:

Homework Statement


a and b are functions of z:

a=a(z); b=b(z)

I want to calculate the second order partial derivative operator on z
Do you mean that you want to think of z as a function of a and b and find
[tex]\frac{\partial^2 z}{\partial a^2}[/tex]
[tex]\frac{\partial^2 z}{\partial b^2}[/tex]
and
[tex]\frac{\partial^2 z}{\partial a\partial b}[/tex]?

Homework Equations


Using the chain rule:

[tex] \frac{\partial}{\partial z}=\frac{\partial a}{\partial z}\frac{\partial}{\partial a}+\frac{\partial b}{\partial z}\frac{\partial}{\partial b}[/tex]

The Attempt at a Solution


Is it correct?
[tex] \frac{\partial^2}{\partial z^2}=\frac{\partial}{\partial a^2}+2\frac{\partial^2}{\partial a\partial b}+\frac{\partial^2}{\partial b^2}[/tex]
This makes no sense at all. What function are you differentiating?

 

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