I'm learning tensor calculus by myself through lectures and texts, and I'm presented with the problem of finding the first and second order derivatives of a scalar function of three variables that vary with two parameters. I label the function $f(x^i(\mu^{\alpha}))$ where $i$ runs from one to three and $\alpha$ runs from one to two, setting the $x^i$ as the variables that change with the parameters $\mu^{\alpha}$. Geometrically this represents a surface in $\mathbb{R}^3$. I also set forth the convention for this problem will be to take Latin indices to run from one to three ( coordinates ) and Greek indicies to run from one to two ( parameters ).
I first begin by finding the general form of the first derivatives of $f$ with respect to arbitrary coordinates. Brute forcing them with multivariate calculus I get: $$ \frac{\partial f}{\partial \mu^1} = \frac{\partial f}{\partial x^1} \frac{\partial x^1}{\partial \mu^1} + \frac{\partial f}{\partial x^2} \frac{\partial x^2}{\partial \mu^1} + \frac{\partial f}{\partial x^3} \frac{\partial x^3}{\partial \mu^1} $$
The same follows for differentiation w.r.t. the second parameter as well. The pattern makes itself clear and allows me to combine both of those derivatives into the single statement:
$$\frac{\partial f}{\partial \mu^{\alpha}} = \frac{\partial f}{\partial x^i} \frac{\partial x^i}{\partial \mu^{\alpha}}$$
following the Einstein Summation Convention. As for the second derivative, I attempted to carry on the tensor notation and not derive results from multivariable calculus. Beginning with a first rank, covariant tensor, I take the derivative of the previous expression with respect to the parameters again, but using a different index to include all possible second order derivative combos. In symbols: $$
\frac{\partial}{\partial \mu^{\beta}}( \frac{\partial f}{\partial \mu^{\alpha}}) = \frac{\partial}{\partial \mu^{\beta}}(\frac{\partial f}{\partial x^i} \frac{\partial x^i}{\partial \mu^{\alpha}}) $$
Applying the chain rule–if done correctly–I get $$ \frac{\partial^2 f}{\partial \mu^{\alpha} \partial \mu^{\beta}} = \frac{\partial^2 f}{\partial x^i \partial x^j}\frac{\partial x^i}{\partial \mu^{\beta}}\frac{\partial x^j}{\partial \mu^{\alpha}} + \frac{\partial f}{\partial x^i} \frac{\partial^2 x^i}{\partial \mu^{\alpha} \partial \mu^{\beta}} $$ a second order covariant tensor. Is this correct?
Best Answer
Check Kryszig's Differential Geometry (Dover, 1959) on p. 156. Though the deduction process is not indicated, a generic expression in index notation is reported which could be useful to corroborate your procedure.