implicit_function_derivatives

$\bold{f}(\bold{x}, \bold{y}) = \bold{0}$ $\bold{f}$ $\bold{y}$ $n$ $\bold{x}$ $m$ $\bold{x}$ $\bold{f}(\bold{x}, \bold{y}) = \bold{0}$ $\bold{y}$ $\bold{y}$ implicit $\bold{x}$ .

$\bold{y} \coloneqq \bold{y}(\bold{x})$ .

$\bold{y}(\bold{x})$ ?

To start, we can differentiate the system of equations to obtain

f (x, y) = 0 ⟹ d f = 0

$\displaystyle \frac{d\bold{y}}{d\bold{x}}$

\begin{aligned} 0 & = d f_{i} \\ 0 & = \frac{\partial f_{i}}{\partial x_{k}} d x_{k} + \frac{\partial f_{i}}{\partial y_{j}} d y_{j} \\ \frac{\partial f_{i}}{\partial y_{j}} d y_{j} & = - \frac{\partial f_{i}}{\partial x_{k}} d x_{k} \\ \frac{\partial f_{i}}{\partial y_{j}} \frac{d y_{j}}{d x_{k}} & = - \frac{\partial f_{i}}{\partial x_{k}} \\ \frac{\partial y_{l}}{\partial f_{i}} \frac{\partial f_{i}}{\partial y_{j}} \frac{d y_{j}}{d x_{k}} & = - \frac{\partial y_{l}}{\partial f_{i}} \frac{\partial f_{i}}{\partial x_{k}}, \frac{\partial y_{i}}{\partial f_{j}} : = [\frac{\partial f}{\partial y}]_{i j}^{- 1} \\ δ_{l j} \frac{d y_{j}}{d x_{k}} & = - \frac{\partial y_{l}}{\partial f_{i}} \frac{\partial f_{i}}{\partial x_{k}} \\ \frac{d y_{l}}{d x_{k}} = - \frac{\partial y_{l}}{\partial f_{i}} \frac{\partial f_{i}}{\partial x_{k}} \end{aligned}

We can follow a similar process to get the Hessian, but it is more involved. Start by differentiating our gradient expression above

\begin{aligned} \frac{d^{2} y_{l}}{d x_{k} d x_{m}} & = \frac{d}{d x_{m}} (\frac{d y_{l}}{d x_{k}}) \\ = - \frac{d}{d x_{m}} (\frac{\partial y_{l}}{\partial f_{i}} \frac{\partial f_{i}}{\partial x_{k}}) \\ = - \frac{d}{d x_{m}} (\frac{\partial y_{l}}{\partial f_{i}}) \frac{\partial f_{i}}{\partial x_{k}} - \frac{\partial y_{l}}{\partial f_{i}} \frac{d}{d x_{m}} (\frac{\partial f_{i}}{\partial x_{k}}) \\ = - (- \frac{\partial y_{l}}{\partial f_{p}} \frac{d}{d x_{m}} (\frac{\partial f_{p}}{\partial y_{q}}) \frac{\partial y_{q}}{\partial f_{i}}) \frac{\partial f_{i}}{\partial x_{k}} - \frac{\partial y_{l}}{\partial f_{i}} (\frac{\partial^{2} f_{i}}{\partial x_{k} \partial x_{m}} + \frac{\partial^{2} f_{i}}{\partial x_{k} \partial y_{j}} \frac{d y_{j}}{d x_{m}}) \\ = \frac{\partial y_{l}}{\partial f_{p}} (\frac{\partial^{2} f_{p}}{\partial y_{q} \partial x_{m}} + \frac{\partial^{2} f_{p}}{\partial y_{q} \partial y_{j}} \frac{d y_{j}}{d x_{m}}) \frac{\partial y_{q}}{\partial f_{i}} \frac{\partial f_{i}}{\partial x_{k}} - \frac{\partial y_{l}}{\partial f_{i}} (\frac{\partial^{2} f_{i}}{\partial x_{k} \partial x_{m}} + \frac{\partial^{2} f_{i}}{\partial x_{k} \partial y_{j}} \frac{d y_{j}}{d x_{m}}) \\ = - \frac{\partial y_{l}}{\partial f_{p}} (\frac{\partial^{2} f_{p}}{\partial y_{q} \partial x_{m}} + \frac{\partial^{2} f_{p}}{\partial y_{q} \partial y_{j}} \frac{d y_{j}}{d x_{m}}) \frac{d y_{q}}{d x_{k}} - \frac{\partial y_{l}}{\partial f_{i}} (\frac{\partial^{2} f_{i}}{\partial x_{k} \partial x_{m}} + \frac{\partial^{2} f_{i}}{\partial x_{k} \partial y_{j}} \frac{d y_{j}}{d x_{m}}) \\ \frac{d^{2} y_{l}}{d x_{k} d x_{m}} = - \frac{\partial y_{l}}{\partial f_{p}} (\frac{\partial^{2} f_{p}}{\partial x_{k} \partial x_{m}} + \frac{\partial^{2} f_{p}}{\partial x_{k} \partial y_{j}} \frac{d y_{j}}{d x_{m}} + \frac{\partial^{2} f_{p}}{\partial y_{q} \partial x_{m}} \frac{d y_{q}}{d x_{k}} + \frac{\partial^{2} f_{p}}{\partial y_{q} \partial y_{j}} \frac{d y_{q}}{d x_{k}} \frac{d y_{j}}{d x_{m}}) \end{aligned}

Here is a mathematica notebook verifying the correctness of these expressions for scalar and vector-valued functions.