4.4 Higher Order Derivatives
Definition 4.4.1. Let \(I\subseteq \mathbb {R}\) be an open interval and let \(f: I \longrightarrow \mathbb {R}\) be differentiable at every point of \(I\). If \(c\in I\) and \(f'\) is itself differentiable at \(c\), then \(f\) is said to have a second order derivative at \(c\), denoted \(f''(c)\): \[f''(c) = \lim _{x\rightarrow c} \frac {f'(x) - f'(c)}{x - c}.\] We generalize this to define the derivative of order \(n\).
Definition 4.4.2. Let \(I\subseteq \mathbb {R}\) be an open interval, \(c\in I\) and \(n\geq 2\). Suppose that \(f: I\longrightarrow \mathbb {R}\) has an \((n - 1)^{\text {th}}\) order derivative \(f^{(n-1)}\) at every point of \(I\). If \(f^{(n-1)}\) is differentiable at \(c\), its derivative is called the \(n^{\text {th}}\) order derivative of \(f\) at \(c\) and is denoted \(f^{(n)}(c)\): \[f^{(n)}(c) = \lim _{x\rightarrow c} \frac {f^{(n - 1)}(x) - f^{(n-1)}(c)}{x - c}.\]
Definition 4.4.3. If \(f\) has an \(n^{\text {th}}\) order derivative at every point of \(I\), we say that \(f\) is \(n\) times differentiable on \(I\).
Example 4.4.4. Find the \(n^{\text {th}}\) derivatives of \(f(x) = \frac {1}{1-x}\) (\(x\neq 1\)) and of \(g(x) = \sin x\).
Solution. We have \(f'(x) = (1-x)^{-2}\), \(f''(x) = 2(1-x)^{-3}\), \(f'''(x) = 6(1-x)^{-4}\), and by induction \(f^{(n)}(x) = n!\,(1-x)^{-(n+1)}\): if this holds for \(n\), differentiating gives \(n!(n+1)(1-x)^{-(n+2)} = (n+1)!(1-x)^{-(n+2)}\). In particular \(f^{(n)}(0) = n!\).
For the sine, \(g' = \cos \), \(g'' = -\sin \), \(g''' = -\cos \), \(g^{(4)} = \sin \), so the derivatives repeat with period \(4\). Equivalently \(g^{(n)}(x) = \sin \left (x + \frac {n\pi }{2}\right )\).
Example 4.4.5. Let \(g(x) = x^2\sin \frac {1}{x}\) for \(x\neq 0\) and \(g(0) = 0\). Show that \(g\) is differentiable on \(\mathbb {R}\) but that \(g''(0)\) does not exist.
Solution. We showed earlier that \(g'(0) = 0\) and \(g'(x) = 2x\sin \frac {1}{x} - \cos \frac {1}{x}\) for \(x\neq 0\). Then \[\frac {g'(x) - g'(0)}{x - 0} = 2\sin \frac {1}{x} - \frac {1}{x}\cos \frac {1}{x},\] which is unbounded near \(0\) (at \(x = \frac {1}{2k\pi }\) it equals \(-2k\pi \)). So \(g'\) is not differentiable at \(0\). In fact \(g'\) is not even continuous at \(0\). So “differentiable” and “continuously differentiable” are genuinely different conditions.
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