We have learned the meaning of the first derivative of a function. Now we want to know what the second, the third, and the n-th derivatives of a function are defined and how we can calculate them. 

Read about the importance of the second derivative in physics

If $s(t)$ is the position of an object moving on a straight line, then the derivative of s is the velocity of the object $v(t)=s'(t)$. The derivative of the velocity is the acceleration of the object $a(t)$. So $a(t)=v'(t)$ or $a(t)=(s’)'(t)$, which is often written simply as $a(t)=s^{\prime\prime}(t)$.We say the acceleration is the second derivative of the position. In physics, acceleration plays an important role as it appears in Newton’s second law $F=ma$.


In general, if we take the derivative of $y=f(x)$, we obtain a new function $f’$ (also denoted by $y’$ or $dy/dx$). We can take the derivative of $f’$ and obtain another function called the second derivative of $f$ (or $y$). The second derivative of $f(x)$ is denoted by \(f^{\prime\prime}(x)\) or \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\), which is commonly abbreviated to \(\dfrac{d^{2}y}{dx^{2}}.\) Thus $f^{\prime\prime}=(f’)^\prime$ or
\[\bbox[#F2F2F2,5px,border:2px solid black]{f^{\prime\prime}(x)=\lim_{\Delta x\to0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x}.\tag{a}}\]
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 \[y’=f'(x)=\frac{dy}{dx}=\frac{df}{dx}=3x^{2}-10x+3,\] and the second derivative of \(f\) is the derivative of \(f'(x)\):

In a similar fashion, we can define the third derivative as the derivative of the second derivative. It is denoted by
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:

  • If $f^{(n)}(x_0)$ exists, then it is said that $f$ is $n$-times differentiable at $x_0$.
Example 1

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


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

Therefore, $y^{(4)}=y=\sin x$.

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 \(f^{(4)}(x)\).


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

Example 3

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


Because the second derivatives of $y=x^3$ and $y=ax^2+bx+c$ exist everywhere, no matter what $a, b$, and $c$ are $f^{\prime\prime}(x)$ exists everywhere, except possibly at $x=1$ where the formula of $f$ changes. So if we want to make $f$ twice differentiable everywhere, we have to choose $a, b$, $c$ such that $f^{\prime\prime}(1)$ exists.

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) \(f_{+}^{\prime\prime}(1)=f_{-}^{\prime\prime}(1)\).


(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

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


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 the Section on Theorems for Calculating Limits  for more information.]

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

Now we can simply plug the formula for \(\frac{d}{dx}\sin\frac{1}{x}\) in (i)
\[\begin{align} f'(x) & =2x\sin\frac{1}{x}+x^{2}\left(-\frac{1}{x^{2}}\right)\left(\cos\frac{1}{x}\right)\\ & =2x\sin\frac{1}{x}-\cos\frac{1}{x}\quad(x\neq0)\end{align}\]
\[f'(x)=\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} & \text{if }x\neq0\\ 0 & \text{if }x=0 \end{cases}.\]
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.