I'm trying to understand a derivation of Bernoilli's Equation and I'm having a hard time understanding the math behind this indentity
$$ \frac{1}{\rho} \nabla p = \nabla \int \frac{\textrm{d}p}{\rho}.$$
I should say that this is for a barotropic flow so $p=p(\rho).$ Trying to remove the physics jargon, I am asking is it true in general that
$$ \frac{1}{g(f)} \nabla f(g) = \nabla \int \frac{\textrm{d}f}{g}.$$
I've tried showing the LHS is equal by doing differentiation under the integral but I'm a bit lost, from the notation and I suppose math. My guess is the bounds are meant to be from $0$ to $p$. Then I get the first term back, but now I'm stuck with an integral I wish to say was zero, but don't know how.
$$ \nabla \int_0^p \frac{\textrm{d}p}{\rho} = \frac{1}{\rho} \nabla p + \int_0^p \nabla \left(\frac{1}{\rho}\right) \textrm{d}p.$$
edit: I misinterpreted it in a not jargon way, and now understand how to solve. It should have been that g is also a function of f, since f is only a function of g.
Let $F(f) = \int \frac{\textrm{d}f}{g},$ so note that $\partial_f F = \frac{1}{g}.$ Then
$$ \nabla F(f) = \frac{\partial F}{\partial f} \nabla f = \frac{1}{g} \nabla f.$$
So in my problem
$$\nabla \int \frac{\textrm{d}p}{\rho} = \frac{\partial}{\partial p} \left(\int \frac{\textrm{d}p}{\rho}\right) \nabla p = \frac{1}{\rho} \nabla p.$$