# Proving the Second Derivative Test using directional derivatives

This is a proof which only uses directional derivatives.

We want to prove the Second Derivative Test.

Suppose the second partial derivatives of $f$ are continuous on a disk with center (a, b), and suppose that $f_x(a,b)=f_y(a,b)=0$. Let

$H=\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2$

(a) If $H>0, f_{xx}(a,b)>0$, then $f(a,b)$ is a local minimum.

(b) If $H>0, f_{xx}(a,b)<0$, then $f(a,b)$ is a local maximum.

(c) If $H<0$, then $f(a,b)$ is a saddle point.

(d) If $H=0$, inconclusive.

Let $u=(h, k)$, which is a unit vector.

\begin{align} D^2_uf(x,y)&=D_u[D_uf(x,y)]\\ &=D_u[\nabla f\cdot u]\\ &=(f_{xx}h+f_{yx}k)h+(f_{xy}h+f_{yy}k)k \end{align}

Since the second partial derivatives are continuous, $f_{xy}=f_{yx}$

\begin{align} D^2_uf(x,y)&=f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\\ &=k^2(f_{xx}(\frac{h}{k})^2+2f_{xy}\frac{h}{k}+f_{yy}) \end{align}

Let $z=\frac{h}{k}$, $g(z)=f_{xx}z^2+2f_{xy}z+f_{yy}$

Use the quadratic function above to find the range of solutions of $D^2_uf(x,y)$.

Let $d=(2f_{xy})^2-4f_{xx}f_{yy}=4(f_{xy}^2-f_{xx}f_{yy})=-4H$

(1) If $d<0$, means that $D^2_uf(x,y)$ can only be either positive or negative.

We can know that the point $(a,b)$ must be a local minimum or a local maximum.

\begin{align} D^2_uf(x,y)&=f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\\ &=f_{xx}(h^2+2\frac{f_{xy}}{f_{xx}}hk+\frac{f_{yy}}{f_{xx}}k^2)\\ &=f_{xx}[(h+\frac{f_{xy}}{f_{xx}}k)^2+(f_{xx}f_{yy}-f_{xy}^2)k^2]\\ &=f_{xx}[(h+\frac{f_{xy}}{f_{xx}}k)^2+Hk^2] \end{align}

We know that $H>0$, so,

if $f_{xx}>0$, the equation above will always be positive, which means a local minimum at $(a,b)$.

Otherwise if $f_{xx}<0$, it will always be negative, which means a local maximum at $(a,b)$.

(2) If $d>0$, means that $D^2_uf(x,y)$ can be both positive, zero or negative, so

$H<0$ tells us that the function has a saddle point at $(a,b)$.

(3) If $d>0$, means that $D^2_uf(x,y)$ can only be either positive and zero or negative and zero, so

$H=0$ cannot tell that whether $(a,b)$ is a local minimum, local maximum or a saddle point.

Note: Posted at The “second derivative test” for $f(x,y)$.

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