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For example, the piecewise function

$ f(x) = \begin{cases} \sqrt{1 - (x + 1)^2} &-2 \leq x \leq 0 \\ -\sqrt{1 - (x - 1)^2} &0 \leq x \leq 2 \end{cases} $

will, at $f(0)$, give $f'(0) = $ undefined (vertical tangent). Once deriving I can prove this algebraically, there is a zero in the denominator. It looked like an inflection point, so I wondered if $f''(x)$ would equal zero or undefined. After taking the (painstakingly ugly) second derivative, I ran into another zero in the denominator error, so $f''(0)$ is also undefined. Is this a general rule that if $f'(x)$ is undefined $f''(x)$ will also be undefined?

Thanks :)

Avritt Rohwer
  • 23

3 Answers3

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Remember the definition of the derivative: $$ f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}. $$ For this definition to make any sense, $f(x)$ must be defined. In your example, since $f'(0)$ is undefined, neither is $f''(0)$.

William Stagner
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mathematician has extended the concept of derivatives into left and right derivatives to deal the case you are facing. As long %x =0$ is negligible for you, you can replace it with left and right derivatives at that point.

See Wiki: Left and Right derivatives

runaround
  • 2,210
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Think of this with the fact that

Differentiable implies continuous

That is, if $g$ is differentiable, then $g$ is continuous. Apply this to $g=f'$.

If $f'$ is differentiable at $x$, that is $f''(x)$ exists, then $f'$ is continuous at $x$. This implies $f'(x)$ exists.