Ok here it is. This example is from Kaplan's book - the detail is provided by me. Given a system of two functions:
<br />
F \equiv ( x^2 + y^2 + u^2 + v^2 - 1)<br />
<br />
G \equiv ( x^2 + 2y^2 - u^2 + v^2 - 1)<br />
Where u and v depend on x and y:
<br />
u = f(x,y)<br />
<br />
v = g(x,y)<br />
find
<br />
\frac{\partial^2 x}{\partial u^2}<br />
To find the first derivative, the method of Jacobians can be used. I am not going to provide the details, but the result is:
<br />
\frac{\partial x}{\partial u} = - \frac{3u}{x}
Finding the second derivative depends on the fact that the differential operator can be chained:
\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \frac{\partial x}{\partial u}
So
\frac{\partial^2 x}{\partial u^2} = \frac{\partial}{\partial u} \frac{\partial x}{\partial u} = \frac{\partial}{\partial u} (- \frac{3u}{x})
Considering the function as a product of two functions and applying the product rule for derivatives produces:
\frac{\partial}{\partial u} [(- 3u)(\frac{1}{x})] = [\frac{\partial}{\partial u} (- 3u)](\frac{1}{x}) - 3u[\frac{\partial}{\partial u} (\frac{1}{x})]
After some manipulation:
- \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})
And finally, the answer:
- \frac{3}{x} - \frac{9y^2}{x^3}
