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Why don't they appear to work?Simply repeating the methods used to calculate the 1st derivatives doesn't appear to work. Any information would be appreciated.

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Ok here it is. This example is from Kaplan's book - the detail is provided by me. Given a system of two functions:

[tex]

F \equiv ( x^2 + y^2 + u^2 + v^2 - 1)

[/tex]

[tex]

G \equiv ( x^2 + 2y^2 - u^2 + v^2 - 1)

[/tex]

Where u and v depend on x and y:

[tex]

u = f(x,y)

[/tex]

[tex]

v = g(x,y)

[/tex]

find

[tex]

\frac{\partial^2 x}{\partial u^2}

[/tex]

To find the first derivative, the method of Jacobians can be used. I am not going to provide the details, but the result is:

[tex]

\frac{\partial x}{\partial u} = - \frac{3u}{x}[/tex]

Finding the second derivative depends on the fact that the differential operator can be chained:

[tex]\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \frac{\partial x}{\partial u}[/tex]

So

[tex]\frac{\partial^2 x}{\partial u^2} = \frac{\partial}{\partial u} \frac{\partial x}{\partial u} = \frac{\partial}{\partial u} (- \frac{3u}{x})[/tex]

Considering the function as a product of two functions and applying the product rule for derivatives produces:

[tex]\frac{\partial}{\partial u} [(- 3u)(\frac{1}{x})] = [\frac{\partial}{\partial u} (- 3u)](\frac{1}{x}) - 3u[\frac{\partial}{\partial u} (\frac{1}{x})] [/tex]

After some manipulation:

[tex]- \frac{3}{x} - 3u[\frac{\partial x}{\partial u}(\frac{\partial}{\partial x}\frac{1}{x})] = - \frac{3}{x} - 3u(- \frac{3u}{x})(- \frac{1}{x^2})[/tex]

And finally, the answer:

[tex]- \frac{3}{x} - \frac{9y^2}{x^3}[/tex]

[tex]

F \equiv ( x^2 + y^2 + u^2 + v^2 - 1)

[/tex]

[tex]

G \equiv ( x^2 + 2y^2 - u^2 + v^2 - 1)

[/tex]

Where u and v depend on x and y:

[tex]

u = f(x,y)

[/tex]

[tex]

v = g(x,y)

[/tex]

find

[tex]

\frac{\partial^2 x}{\partial u^2}

[/tex]

To find the first derivative, the method of Jacobians can be used. I am not going to provide the details, but the result is:

[tex]

\frac{\partial x}{\partial u} = - \frac{3u}{x}[/tex]

Finding the second derivative depends on the fact that the differential operator can be chained:

[tex]\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \frac{\partial x}{\partial u}[/tex]

So

[tex]\frac{\partial^2 x}{\partial u^2} = \frac{\partial}{\partial u} \frac{\partial x}{\partial u} = \frac{\partial}{\partial u} (- \frac{3u}{x})[/tex]

Considering the function as a product of two functions and applying the product rule for derivatives produces:

[tex]\frac{\partial}{\partial u} [(- 3u)(\frac{1}{x})] = [\frac{\partial}{\partial u} (- 3u)](\frac{1}{x}) - 3u[\frac{\partial}{\partial u} (\frac{1}{x})] [/tex]

After some manipulation:

[tex]- \frac{3}{x} - 3u[\frac{\partial x}{\partial u}(\frac{\partial}{\partial x}\frac{1}{x})] = - \frac{3}{x} - 3u(- \frac{3u}{x})(- \frac{1}{x^2})[/tex]

And finally, the answer:

[tex]- \frac{3}{x} - \frac{9y^2}{x^3}[/tex]

Last edited:

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Are u using only the first equation (the one for F), or are u using the second (for G) as well?

Daniel.

Daniel.

Last edited:

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[tex]\frac{\partial x}{\partial u} = - \frac{3u}{x}[/tex].

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It's actually in reversed orderrick1138 said:Finding the second derivative depends on the fact that the differential operator can be chained:

[tex]\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \frac{\partial x}{\partial u}[/tex]

[tex] \frac{\partial}{\partial u} = \frac{\partial x}{\partial u} \frac{\partial}{\partial x}[/tex]

Daniel.

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