Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 3

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

Use implicit differentiation on $F(x,y)=x{y}^3-y{x}^3-6=0$. Along the curve,At $(1,2)$, $F_x=2$ and $F_y=11\ne 0$, so the slope is $-2/11$. The tangent line is

For $F(x,y)=e^{xy}+y\ln x-x^2$, the real domain has $x>0$. Its partial derivatives are $F_x=y{e}^{xy}+y/x-2x$ and $F_y=x{e}^{xy}+\ln x$. At $(1,0)$ they equal $-2$ and $1$, respectively. Thus implicit differentiation gives $y^{\prime }=2$, and the tangent line is