first and second partial derivative of integral of a vector norm

230 Views Asked by At

I'm not very familiar with the Leibniz rule or chain rule of differentiation. Could anyone check whether my calculations are correct?

I need to find the first and second partial derivative of the following function:

$$f(\vec x) = \int_{-1}^{1} \int_{-1}^{1} \lvert\phi(\vec x(s, t))\rvert\,ds\,dt$$ $\lvert\cdot\rvert$ is the Euclidean norm, and $\vec{x} = (x_1, x_2, x_3)$ , $x_i =x_i(s,t)$ and $\phi(\vec{x}) = \{\phi_1(\vec{x}) , \phi_2(\vec{x}), \phi_3(\vec{x})\}$.

My calculations: $$ \frac {\partial f(\vec x)}{\partial x_i} = \int_{-1}^{1} \int_{-1}^{1} \ \frac{1}{\lvert\phi(\vec x)\rvert} \left[ \phi(\vec x)\ {\frac{\partial \phi(\vec x)}{\partial x_i}}^T \right] \,ds\,dt$$

$$ \frac{\partial f(\vec x)}{\partial x_i \partial x_j} = \int_{-1}^{1} \int_{-1}^{1} \left[ \frac{-1}{2}\ \frac{1}{\lvert\phi(x)\rvert^3 } \left( \phi(\vec x)\ {\frac{\partial \phi(\vec x)}{\partial x_j}}^T \right) \left( \phi(\vec x)\ {\frac{\partial \phi(\vec x)}{\partial x_i}}^T \right) \right] + \frac{1}{\lvert \phi(\vec x)\rvert} \left[ \frac {\partial \phi(\vec x)}{\partial x_j} \ {\frac{\partial \phi(\vec x)}{\partial x_i}}^T+ \phi(\vec x)\ {\frac{\partial \phi(\vec x)}{\partial x_i \partial x_j}}^T \right]ds\,dt$$