Because \(f'(x)\) (or \(dy/dx\)) is in general a function of \(x\), it may be differentiated with respect to \(x\). The result is called the second derivative of \(f\) (or \(y\)) with respect to \(x\) and is denoted by \(f^{\prime\prime}(x)\) or \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\), which is commonly abbreviated into \(\dfrac{d^{2}y}{dx^{2}}.\) Thus
\[\bbox[#F2F2F2,5px,border:2px solid black]{f^{\prime\prime}(x)=\lim_{\Delta x\to0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x}.\tag{a}}\]
The second derivative is also indicated by \(y^{\prime\prime}\) or
\(\dfrac{d^{2}f}{dx^{2}}.\)

For example, if \(f(x)=x^{3}-5x^{2}+3x-1\), then the first derivative is \[\frac{dy}{dx}=y’=f'(x)=3x^{2}-10x+3,\] and the second derivative of \(f\) is the derivative of \(f'(x)\):
\[\frac{d^{2}y}{dx^{2}}=y^{\prime\prime}=f^{\prime\prime}(x)=6x-10.\]

In a similar fashion, we can define the third derivative as the derivative of the second derivative. It is denoted by
\[y^{\prime\prime\prime}=f^{\prime\prime\prime}(x)=\frac{d^{3}y}{dx^{4}}=\frac{d^{4}f}{dx^{4}};\]
the fourth derivative is the derivative of the third derivative, and is denoted by
\[y^{(4)}=f^{(4)}(x)=\frac{d^{4}y}{dx^{4}}=\frac{d^{4}f}{dx^{4}},\] and so on. In general, the \(n\)-th derivative of \(y=f(x)\) is indicated by one of the following symbols:
\[y^{(n)}=f^{(n)}(x)=\frac{d^{n}y}{dx^{n}}=\frac{d^{n}f}{dx^{n}}.\]

Example 1

If \(y=\sin x\), find \(y^{(4)}.\)

Solution

\[\begin{align} y & =\sin x\Rightarrow y’=\cos x\Rightarrow y^{\prime\prime}=\frac{d}{dx}\cos x=-\sin x\\ \Rightarrow y^{\prime\prime\prime} & =\frac{d}{dx}(-\sin x)=-\cos x\Rightarrow y^{(4)}=\dfrac{d}{dx}(-\cos x)=-(-\sin x)=\sin x.\end{align}\]

Example 2

For \(f(x)=-x^{5}+3x^{3}+2x^{2}-1\), find \(f'(x),f^{\prime\prime}(x),f^{\prime\prime\prime}(x)\) and \(f^{(4)}(x)\).

Solution

\[\begin{align}f^{\prime}(x)  &=-5x^{4}+9x^{2}+4x\\ f^{\prime\prime}(x)  &=-20x^{3}+18x+4\\ f^{\prime\prime\prime}(x)  &=-60x^{2}+18\\ f^{(4)}(x)  &=-120x.\end{align}\]

Example 3

Determine \(a,b\), and \(c\) such that \(f^{\prime\prime}(x)\) exists everywhere if
\[f(x)=\begin{cases} x^{3} & \text{when }x\leq1\\ ax^{2}+bx+c & \text{when }x>1 \end{cases}.\]

Solution

We have three unknowns: \(a,b,c\) and hence we need three equations. For \(f\) to have a second derivative at \(x=1\), we need:
(1) \(f\) to be continuous at \(x=1\),
(2) \(f\) to have a derivative at \(x=1\) or \(f’_{-}(1)=f’_{+}(1)\), and
(3) \(f_{+}^{\prime\prime}(1)=f_{-}^{\prime\prime}(1)\).
Now
(1) The continuity of \(f\) at \(x=1\) implies \[f(1)=1^{3}=a\cdot1^{2}+b\cdot1+c.\tag{i}\] (2) \(f\) has a derivative at \(x=1\). Thus
\[\begin{align} f’_{-}(1) & =f’_{+}(1)\\ \left.3x^{2}\right|_{x=1} & =\left.2ax+b\right|_{x=1}\\ 3 & =2a+b.\tag{ii}\end{align}\]
(3) \(f\) has a second derivative at \(x=1\). Thus
\[\begin{align} f^{\prime\prime}_{-}(1) & =f^{\prime\prime}_{+}(1)\\ \left.6x\right|_{x=1} & =\left.2a\right|_{x=1}\\ 6 & =2a\tag{iii }\end{align}\]
From (i), (ii), and (iii), we conclude
\[a=3,\qquad b=-3,\quad\text{and}\quad c=1.\]

  • If a function is differentiable, its derivative is not necessarily differentiable. In other words, from the existence of \(f'(x_{0})\), we cannot infer the existence of \(f^{\prime\prime}(x_{0})\). For instance, see the following example.
Example 4

Let
\[f(x)=\begin{cases} x^{2}\sin\frac{1}{x} & x\neq0\\ 0 & x=0 \end{cases}.\]
Does \(f'(0)\) exist? Is \(f'(x)\) continuous at \(x=0\)?

Solution

To find \(f'(0)\), we need to apply the definition of a derivative directly:
\[\require{cancel}\begin{align} f'(0) & =\lim_{\Delta x\to0}\frac{f(0+\Delta x)-\overset{0}{\cancel{f(0)}}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{(\Delta x)^{2}\sin\dfrac{1}{\Delta x}}{\Delta x}\\ & =\lim_{\Delta x\to0}\left(\Delta x\sin\dfrac{1}{\Delta x}\right)\\ & =0.\end{align}\]
[Recall that \(\lim_{h\to0}h\sin\frac{1}{h}=0\). See Section 4.4  for more information]

When \(x\neq0\), we can find \(f'(x)\) by using differentiation rules:
\[f(x)=\underbrace{x^{2}}_{u}\underbrace{\sin\frac{1}{x}}_{v}\]
\[f'(x)=\underbrace{2x}_{u’}\underbrace{\sin\frac{1}{x}}_{v}+\underbrace{x^{2}}_{u}\underbrace{\frac{d}{dx}\sin\frac{1}{x}}_{v’}\tag{i}\]
To find \(\frac{d}{dx}\sin\frac{1}{x}\) let \(w=\frac{1}{x}\)
\[\begin{align} \frac{d}{dx}\sin\frac{1}{x} & =\frac{d}{dx}\sin w\\ & =\frac{d}{dw}(\sin w)\times\frac{dw}{dx}\\ & =(\cos w)\left(-\frac{1}{x^{2}}\right)\\ & =\left(\cos\frac{1}{x}\right)\left(-\frac{1}{x^{2}}\right)\end{align}\]
[\(\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}x^{-1}=-x^{-2}=-\frac{1}{x^{2}}\)]

Now we can simply plug the formula for \(\frac{d}{dx}\sin\frac{1}{x}\) in (i)
\[\begin{align} f'(x) & =2x\sin\frac{1}{x}+x^{2}\left(-\frac{1}{x^{2}}\right)\left(\cos\frac{1}{x}\right)\\ & =2x\sin\frac{1}{x}-\cos\frac{1}{x}\quad(x\neq0)\end{align}\]
Therefore,
\[f'(x)=\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} & \text{if }x\neq0\\ 0 & \text{if }x=0 \end{cases}.\]
Because \(\cos(1/x)\) moves up and down so quickly as \(x\to0\), it does not approach a number, and \(\lim_{x\to0}\cos(1/x)\) does not exists. Thus
\[\begin{align} \lim_{x\to0}f'(x) & =\lim_{x\to0}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)\\ & =2\lim_{x\to0}x\sin\frac{1}{x}-\lim_{x\to0}\cos\frac{1}{x}\\ & =2(0)-DNE\end{align}\]
does not exists, and consequently \(f’\) is not continuous at \(x=0\).

  • In the above example, \(f\) is differentiable (= \(f'(x)\) exists) everywhere. But because \(f'(x)\) is not continuous at \(x=0\), \(f^{\prime\prime}(0)\) does not exist.
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