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 $x_1 \in \mathbb{R}$


$\frac{\partial u(x_1, f(x_1))}{\partial x_1} + \frac{\partial u(x_1, f(x_1))}{\partial x_2}f'(x_1) = 0$.

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 $\frac{du(x,y)}{dx} = u_1\frac{dx}{dx} + u_2 \frac{dy}{dx}$.


As long as $u$ and $f$ are real-valued and $x_1\in\mathbb R$, this should be okay (as opposed to, say, vector fields and a vector, respectively).

  • $\begingroup$ Yes, sorry I should have clarified. $u : \mathbb{R^{2}_+ \rightarrow \mathbb{R_+}}, f : \mathbb{R_+} \rightarrow \mathbb{R_+}$ and $x_1 \in \mathbb{R_+}$. $\endgroup$ – student_t Sep 24 '17 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.