Past exam of the mathematical methods II course of the University of Oxford / 2013 resit / 4 / Solution

Substitute the path into the height before differentiating: $x^2=1+t$, $y=t-1$, soThe ordinary chain rule immediately yieldsThe multivariable calculation gives the same result: $h_x=-2x/(x^2+2y^4+1{)}^2$, $h_y=-8y^3/(x^2+2y^4+1{)}^2$, and $\dot{h}=h_x\dot{x}+h_y\dot{y}$ with $\dot{x}=1/(2\sqrt{1+t})$ and $\dot{y}=1$. The real path has $t\ge -1$; this chain-rule calculation applies for $t>-1$, and the simplified height has a finite right derivative at the endpoint even though $\dot{x}$ there is singular.