I want to find an expression for $f''(x_1)$ from the equation below. Note that the partial derivative of the first slot and second slot of $u(x_1, f(x_1))$ are denoted $u_1, u_2$. Also, for clarification:$u : \mathbb{R^2 \rightarrow \mathbb{R}}, f : \mathbb{R} \rightarrow \mathbb{R}$ and $x_1 \in \mathbb{R}$
Given:
$\frac{\partial u(x_1, f(x_1))}{\partial x_1} + \frac{\partial u(x_1, f(x_1))}{\partial x_2}f'(x_1) = 0$.
So, what I try to do is totally differentiate the expression to get (where I try to use the product rule on the second expression and the chain rule in both):
$\left[\frac{\partial^2 u(x_1, f(x_1))}{\partial x_1 \partial x_1} + \frac{\partial^2 u(x_1, f(x_1))}{\partial x_1 \partial x_2}f'(x_1)\right] + \left[\frac{\partial u(x_1, f(x_1))}{\partial x_2}f''(x_1) + f'(x_1) \left(\frac{\partial^2 u(x_1, f(x_1))}{\partial x_2 \partial x_1} + \frac{\partial^2 u(x_1, f(x_1))}{\partial x_2 \partial x_2}f'(x_1)\right)\right] = 0$
Writing with shorthand:
$u_{11} + u_{12}f'(x_1) + u_2f''(x_1) + f'(x_1)u_{21} + u_{22}(f'(x_1))^2 = 0$
Solving, and I know $u_2 > 0$:
$f''(x_1) = \frac{-u_{11} - u_{12}f'(x_1)-f'(x_1)u_{21} - u_{22}(f'(x_1))^2}{u_2}$
I am really rusty and uncertain about how I took the derivative, can anyone confirm this is correct or have I completely butchered the operation. I am unsure about my application of total differentiation, multi-variable product rule, and chain rule.
My understanding is if you have $u(x, y)$ the total derivative w.r.t $x$ is $\frac{du(x,y)}{dx} = u_1\frac{dx}{dx} + u_2 \frac{dy}{dx}$.
