= Solution
Substitute the path into the height before differentiating: $x^2=1+t$, $y=t-1$, so
$$
h(t)=\frac1{t+2+2(t-1)^4}.
$$
The ordinary <chain rule> immediately yields
$$
\boxed{\frac{dh}{dt}=-\frac{1+8(t-1)^3}{[t+2+2(t-1)^4]^2}.}
$$
The 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 $t\geq-1$; 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.
Back to article page
