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

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.