Derivative of |x^2-9|

[ValeeBee]

I was able to calculate the derivative of |x^2-9|: the result is (2 x (-9 + x^2))/abs(-9 + x^2). However, I have a doubt as to whether it really exists, as it contradicts the axiom that uniqueness of derivatives. Does the derivative really exist? Should I define the derivative with a specific domain and path? The derivative is the red graph and the function to be derived is the blue one.

 

[Dr.Peterson]

That looks to me like it exists, and is unique! What do you see as wrong with it?

Is it the fact that the derivative is discontinuous? That's to be expected, when f has sudden "breaks".

Your derivative is essentially f'(x) = 2x sgn(x^2 - 9), where sgn(u) is the sign of u: +1 when u>0, -1 when u<0, but is undefined (for our purposes) when u=0. It is most clearly defined piecewise. But there is nothing wrong with that.

 

[ValeeBee]

The fact that the function has two values in "y" for x=3 and x=-3 is what made me doubt if (in spite of this) the criterion of uniqueness of the derivatives is fulfilled.

 

[tkhunny]

There are NOT two values of either the function or the derivative for x = 3 or for x = -3. There are NO values for the derivative at those places. It is discontinuous. There's a gap. The derivative fails to exist for x = 3 and for x = -3. There is no "in spite of".

 

[Dr.Peterson]

At x=3 and at x=-3, there is a "left derivative" and a "right derivative", which are not equal. But that means, not that there are two derivatives, but that there is no derivative! When the limit from the left and the limit from the right are not equal, the limit is undefined.

Note that you gave the derivative as (2 x (-9 + x^2))/abs(-9 + x^2); that is a function, and is undefined at those points, right? I imagine you may have given too much attention to the graph instead.

 

[ValeeBee]

Thank you so much!