I need to take the partial derivatives w.r.t. $x^*$ and $y^*$ of the following equation $L(x^*,y^*;x_0,y_0)$:
$L(x^*,y^*;x_0,y_0)=\int_{x_0}^{x^*}\frac{\partial f}{\partial x}\mathrm{d}x+\int_{y_0}^{y^*}\frac{\partial f}{\partial x}\frac{\partial x}{\partial y}\mathrm{d}y.$
The result I'm getting for the partial derivative of $L$ w.r.t. $x^*$ doesn't seem right. First let me expand things a little:
$L(x^*,y^*;x_0,y_0)=f(x^*)-f(x_0)+\frac{\partial f}{\partial x}(x(y^*)-x(y_0))$
(Important to note that $x^=x(y^)$).
I think up to here it's ok. But now when I take the partial derivative...
$\frac{\partial L}{\partial x^*}=\frac{\partial f}{\partial x^*}+\frac{\partial^2 f}{\partial x^{2}}(x^*-x(y_0))+\frac{\partial f}{\partial x}$
I'm confused as to whether the second term $\frac{\partial^2 f}{\partial x^{2}}(x^*-x(y_0))$ should be included in there since, strictly speaking, it is a derivative w.r.t. $x$ and not $x^*$. Or does the fact that $x^*=x(y^*)$ imply that this is a derivitive w.r.t. $x^*$? The last term, on the other hand, is indeed a derivative w.r.t. $x^*$ no matter how you slice it, but then what to do with $\frac{\partial f}{\partial x}$ which might be evaluated at x=anything and thus is effectively a random number thrown into the equation?
My intuition tells me that I should just treat $\frac{\partial^2 f}{\partial x^{2}}$ and $\frac{\partial f}{\partial x}$ here as evaluated at $x^*$, but looking at the actual equations I can't see the justification.
Appreciate any advice in this regard.