Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / F (Multivariable Calculus)

The $j$-th partial derivative is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is $J_f(x)=({\partial }_j{f}_i(x){)}_{i,j}$. When the total derivative exists, all partial derivatives exist and $J_f(x)$ is its matrix in the standard bases; its $j$-th column is $Df(x)e_j$. Existence of partial derivatives alone does not guarantee total differentiability.

For a scalar-valued function, the gradient is the column vectorWhen $f$ is differentiable, its total derivative is the linear functionalThus the Jacobian is the transpose of the gradient in standard Euclidean coordinates.

The chain rule at the specified point isIn standard coordinates this becomesThe matrix dimensions are respectively $k\times n$, $k\times m$, and $m\times n$, matching the composition of the two linear derivative maps.

Put $Q=1+x^2+y^2>0$. Differentiation of $Q^{-1}$ givesThe positive denominator also shows that the function and these derivatives are smooth throughout ${\mathbb{R}}^2$.

At $(-1,0)$ the gradient is $(1/2,0{)}^T$. The total derivative is therefore the linear functional

The derivative of the curve isIts image lies on the unit circle, so $f(g(t))=1/2$ and $\boxed{D(f\circ g)(t)=0}$. Meanwhileand multiplication by $J_g(t)$ gives $-\sin t\cos t/2+\cos t\sin t/2=0$. This verifies the chain rule explicitly.

Because $2|x|\le 1+x^2\le Q$, we have $|f_x(x,y)|=2|x|/Q^2\le 1/Q\le 1$; similarly $|f_y|\le 1$. Apply the one-variable mean value theorem on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The triangle inequality givesThus the function is globally one-Lipschitz for the ${\ell }^1$ metric on the plane.