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I have a question regarding first order derivatives. If we consider the function $f : \mathbb{R}^2\rightarrow\mathbb{R}^3$ with $(x_1, x_2)\mapsto (x^2_1+ x_2, x^3_2, \cos{x_2})$. We have to calculate $f'(0)(a)$ and $f''(0)(a)(b)$ where $a = (a_1, a_2)$ and $b = (b_1, b_2).$

Can some one please tell me how one is supposed to tackle this problem. Thanks.

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Do you mean $f ′(0)(a)(b)$ or simply $f ′(0)(b)$? Only the latter makes sense to me! So I may assume the latter. Just differentiate each co-ordinate w.r.t. x_1, x_2 and write down the derivatives w.r.t. x_i of each function in one row i.e., $$\begin{pmatrix}\frac{df_1}{dx_1}&\frac{df_1}{dx_2}\\\frac{df_2}{dx_1}&\frac{df_2‌​}{dx_2}\\\frac{df_3}{dx_1}&\frac{df_3}{dx_2}\end{pmatrix}$$ Now evaluate at $0$ and left-multiply $a$ and $b$ separately with the resulting $3\times2$ matrix. – Host-website-on-iPage Jul 10 '12 at 10:17
Evaluating at 0 gives the 3 x 2 matrix (0 1;0 0; 0 0). Hence, the derivatives at a is simply (a2 0 0). Is that correct? Also, it is f '' (0) (a) (b). – carla Jul 10 '12 at 10:38

Recall from the theory that the (Fréchet) derivative of a map $F \colon \mathbb{R}^n \to \mathbb{R}^m$ at a given point $x_0$ is the linear map $DF(x_0) \colon \mathbb{R}^n \to \mathbb{R}^m$ represented by the Jacobian matrix evaluated at $x_0$. The jacobian matrix is the $m \times n$ matrix whose entries are the partial derivatives of the $m$ components $F_j$ of $F$ with respect to the $n$ independent variables. If $m=1$, it is the $1 \times n$ matrix $$ \begin{pmatrix} \frac{\partial F}{\partial x_1} &\frac{\partial F}{\partial x_2} &\ldots &\frac{\partial F}{\partial x_n} \end{pmatrix} $$ The second derivative is represented by the Hessian matrix. So you need to compute all the first and second derivatives at $0$ and just construct the matrices.

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