Let $$ \mathbf{f}(\mathbf{x}) = (x_1 + x_2)^2\mathbf{e}_1 + x_1\sin x_2\mathbf{e}_2,\quad \mathbf{x}_0 = \mathbf{e}_1 + \pi\mathbf{e}_2: $$ Find $D_{\mathbf{v}_0 \mathbf{u}_0}\mathbf{f}$ where $\mathbf{u}_0 = (\mathbf{e}_1 + 2\mathbf{e}_2)$ and $\mathbf{v}_0 = (\mathbf{e}_1 - \mathbf{e}_2)$.
How would I approach this problem? Apologies for the poor formatting, this is my first time on the website. I put a better image here. See part (c).
There is a little typo in (c), it should be $D^2_{\mathbf{v}_0 \mathbf{u}_0}\mathbf{f}(\mathbf{x}_0)$. We have that $$D^2\mathbf{f}(\mathbf{x})=\begin{bmatrix} 2\mathbf{e}_1 & 2\mathbf{e}_1+\cos(x_2)\mathbf{e}_2 \\ 2\mathbf{e}_1+\cos(x_2)\mathbf{e}_2 & 2\mathbf{e}_1-x_1\sin(x_2)\mathbf{e}_2 \end{bmatrix},$$ and at the given point $\mathbf{x}_0=\mathbf{e}_1+\pi\mathbf{e}_2$, we find $$D^2\mathbf{f}(\mathbf{x}_0)=\begin{bmatrix} 2\mathbf{e}_1 & 2\mathbf{e}_1-\mathbf{e}_2 \\ 2\mathbf{e}_1-\mathbf{e}_2 & 2\mathbf{e}_1 \end{bmatrix}.$$ It remains to evaluate (see Theorem 7.11, a few pages before, in same the book), $$ \begin{align} D^2_{\mathbf{v}_0 \mathbf{u}_0}\mathbf{f}(\mathbf{x}_0)&=2\mathbf{e}_1 u_1v_1+(2\mathbf{e}_1-\mathbf{e}_2)(u_1v_2+u_2v_1)+2\mathbf{e}_1 u_2v_2\\ &=2\mathbf{e}_1(1\cdot 1)+(2\mathbf{e}_1-\mathbf{e}_2)(1\cdot -1+2\cdot 1)+2\mathbf{e}_1 (2\cdot-1)\\ &=2\mathbf{e}_1+2\mathbf{e}_1-\mathbf{e}_2-4\mathbf{e}_1=-\mathbf{e}_2.\end{align}$$