Divergence of a composite function

Question

I guess my question is completely stupid but I'm lost and I need to clear this up.

I want to take the divergence of a vector field:

$$ \operatorname{div}\begin{bmatrix} 0 \\ 0 \\ f(z,T) \end{bmatrix} = \frac{\partial f}{\partial z}(z,T) . $$

Now what is the problem - I know that $T$ is a function of $z$. In the end, I want to get rid of the $T$ variable. Should I apply the partial derivative on $T$ as well?

To be specific, let's assume that $f(z,T) = z^2\cos T$. What is the divergence of the vector field? This $$2z\cos T(z),$$ or this $$2z\cos T(z) - z^2(\sin T(z)) T'(z). $$

Thank you for any kind of help. If someone could explain this to me I would be very grateful.

Answer 1

It's: $$2z \cos T(z) - z^2 \sin T(z) \cdot T'(z)$$

That is because divergence is defined for a vector function g(x,y,z), which is in your case: $$\vec g(x,y,z) = f(z,T(z)) \hat k$$ This does not involve $x$ or $y$, but still applies to all occurrences of $z$.