# 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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