Let $u:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a smooth function and let $I$ be an interval such that each $a \in I$ is a regular value for $u$.
Now consider the line integral
$$F(a)=\int_{u^{-1}(a)} f(s) ds$$ is there a formula for the derivative of this function $F$?
If anything is unclear, please let me know.
Just a very heuristic computation, but I guess the result would be
\begin{align*} F'(a) &= \int_{u^{-1}(a)} \left( \nabla f \cdot \frac{\nabla u}{|\nabla u|^2} + f \frac{k}{|\nabla u|} \right) \, ds \\ &= \int_{u^{-1}(a)} \left( \nabla f \cdot \frac{\nabla u}{u_x^2+u_y^2} + f \frac{u_{xx}u_y^2 - 2u_{xy}u_x u_y + u_{yy}u_x^2}{(u_x^2+u_y^2)^2} \right) \, ds, \end{align*}
where $k$ is the curvature of the curve $u^{-1}(a)$ measured in the direction of $-\nabla u$.
Example. consider $u = c(x^2+y^2)$ with $c \neq 0$. Then $u^{-1}(a)$ is a circue of radius $\sqrt{a/c}$ centered at the origin $0$ and thus
$$ F(a) = \sqrt{\frac{a}{c}} \int_{0}^{2\pi} f\left( \sqrt{\frac{a}{c}}\cos\theta, \sqrt{\frac{a}{c}}\sin\theta \right) \, d\theta. $$
Direct computation using the above expression leads
\begin{align*} F(a) &= \frac{1}{4ac} \int_{u^{-1}(a)} \nabla f \cdot \nabla u \, ds + \frac{1}{2a} \int_{u^{-1}(a)} f \, ds. \end{align*}
On the other hand, using the formula together with
$$ \frac{1}{u_x^2+u_y^2} = \frac{1}{4c^2(x^2+y^2)}, \qquad \frac{u_{xx}u_y^2 - 2u_{xy}u_x u_y + u_{yy}u_x^2}{(u_x^2+u_y^2)^2} = \frac{1}{2c(x^2+y^2)} $$
leads to the same answer.