Past exam of the options II course of the University of Oxford / 2011 / F / F2

Writing $f=(f_1,\dots ,f_m)$, the $j$th partial derivative at $x$ is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ satisfyingThe Jacobian matrix has entries $J_f(x{)}_{ij}={\partial }_j{f}_i(x)$. If the total derivative exists, its matrix in the standard bases is precisely $J_f(x)$, because substitution of $h=t{e}_j$ identifies every column. Existence of the partial derivatives alone does not in general imply existence of the total derivative.

For a scalar-valued differentiable function, its gradient is the column vectorIt represents the total derivative by the Euclidean inner product:Thus $J_f(x)=\nabla f(x{)}^T$ is a row matrix; the gradient and the total derivative represent the same first-order information but have different types.

The chain rule states that $f\circ g$ is differentiable at $x$ andIn terms of Jacobian matrices,The factors have dimensions $k\times m$ and $m\times n$, respectively, giving the required $k\times n$ Jacobian matrix.

Differentiating the reciprocal of $1+x^2+y^2$ gives the gradient

At $(-1,0)$ the gradient is $(1/2,0{)}^T$. Therefore the total derivative is the linear maprepresented by the Jacobian matrix $\left(\begin{array}{cc}1/2& 0\end{array}\right)$.

The total derivative of $g$ sends a scalar increment $s$ toSince ${\sin }^{2}t+{\cos }^{2}t=1$, the function composition $f\circ g$ is constantly $1/2$, so $D(f\circ g)(t)=0$. Directly, the Jacobian matrix product isThis verifies the chain rule at every $t$.

The partial derivatives satisfy a global bound of $1$ in absolute value. For example,because $2|x|\le 1+x^2$ and $1+x^2\ge 1$; the same argument applies to ${\partial }_{y}f$. Apply the one-dimensional mean value theorem first along a horizontal segment and then along a vertical segment:Thus $f$ is Lipschitz with constant $1$ for the coordinate-sum norm.