# Thread: second derivative along a path

1. ## second derivative along a path

Hello,
My question is :

Show that the function
$f(x,y) = xy$ has a saddle point at $(x,y) = (0,0)$ and find the second derivative along a path through that point has the greatest positive and negative values.

My thoughts:

If I substitute y = x into the function as thus $f(y,y) = x.x = x^2$ and then find the second derivative as thus $d^2f/dx^2 = 2$ this is repeated for the path y = -x as thus $f(y,y) = -x.x = -x^2$with a second derivative of $d^2f/dx^2 = -2$

This is repeated for paths x=y and x=-y giving a derivative of $d^2f/dy^2 = 2$ and $d^2f/dy^2 = -2$ repectively. So the greatest positive and negative values are 2 and -2. At y=0 and x = 0 the function is constant.

Any Help would be greatly appreciated.
Thank you,
Riptorn70.

2. I don't understand your reasoning. You calculated the second derivatives along two different paths and got -2 and 2. What makes you sure that those are the smallest and largest values?

What if we approach (0, 0) along the line y= 10000x? Then $f(x,y)= f(x,10000x)= 10000x^2$. What is the second derivative of that?

Do it more generally. Since the derivative is "local", we can approximate any path through (0, 0) by a straight line: y= mx for some number, m. Then $f(x,y)= f(x, mx)= mx^2$. What is the second derivative of that?

That includes every path except vertical ones- on a line tangent to such a path, at (0, 0), x= 0 so f(x,y)= f(0, y)= 0 and the second derivative is 0.

3. thank you hallofIvy!!