5.9 Higher Derivatives

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.

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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 $a (t) = s^{''} (t)$ .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 $F = m a$ .

In general, if we take the derivative of $y = f (x)$ , we obtain a new function $f^{'}$ (also denoted by $y^{'}$ or $d y / d x$ ). 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 $f^{''} (x)$ or $\frac{d}{d x} (\frac{d y}{d x})$ , which is commonly abbreviated to $\frac{d^{2} y}{d x^{2}} .$ Thus $f^{''} = (f^{'})^{'}$ or
$\begin{matrix} (a) & f^{''} (x) = lim_{Δ x \to 0} \frac{f^{'} (x + Δ x) - f^{'} (x)}{Δ x} . \end{matrix}$
The second derivative is also indicated by $y^{''}$ or $\frac{d^{2} f}{d x^{2}}$ .

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

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

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 $y^{(4)} .$

Solution

$\begin{aligned} y & = \sin x \\ y^{'} & = \cos x \\ y^{''} & = \frac{d}{d x} \cos x = - \sin x \\ y^{'''} & = \frac{d}{d x} (- \sin x) = - \frac{d}{d x} \sin x = - \cos x \\ y^{(4)} & = \frac{d}{d x} (- \cos x) = - \frac{d}{d x} \cos x = - (- \sin x) = \sin x . \end{aligned}$

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

Example 2

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

Solution

$\begin{aligned} f^{'} (x) & = - 5 x^{4} + 9 x^{2} + 4 x \\ f^{''} (x) & = - 20 x^{3} + 18 x + 4 \\ f^{'''} (x) & = - 60 x^{2} + 18 \\ f^{(4)} (x) & = - 120 x . \end{aligned}$

Example 3

Determine $a, b$ , and $c$ such that $f^{''} (x)$ exists everywhere if
$f (x) = {\begin{cases} x^{3} & when x \leq 1 \\ a x^{2} + b x + c & when x > 1 \end{cases} .$

Solution

Because the second derivatives of $y = x^{3}$ and $y = a x^{2} + b x + c$ exist everywhere, no matter what $a, b$ , and $c$ are $f^{''} (x)$ 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 $f^{''} (1)$ 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) $f_{+}^{''} (1) = f_{-}^{''} (1)$ .

Now

(1) The continuity of $f$ at $x = 1$ implies $\begin{matrix} (i) & f (1) = 1^{3} = a \cdot 1^{2} + b \cdot 1 + c . \end{matrix}$ (2) $f$ has a derivative at $x = 1$ . Thus
$\begin{aligned} f_{-}^{'} (1) & = f_{+}^{'} (1) \\ {3 x^{2} |}_{x = 1} & = {2 a x + b |}_{x = 1} \\ (ii) & 3 & = 2 a + b . \end{aligned}$
(3) $f$ has a second derivative at $x = 1$ . Thus
$\begin{aligned} f_{-}^{''} (1) & = f_{+}^{''} (1) \\ {6 x |}_{x = 1} & = {2 a |}_{x = 1} \\ (iii) & 6 & = 2 a \end{aligned}$

From (i), (ii), and (iii), we conclude
$a = 3, b = - 3, and c = 1.$

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 $f^{''} (x_{0})$ . For instance, see the following example.

Example 4

Let
$f (x) = {\begin{cases} x^{2} \sin \frac{1}{x} & x \neq 0 \\ 0 & x = 0 \end{cases} .$
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:
$\begin{aligned} f^{'} (0) & = lim_{Δ x \to 0} \frac{f (0 + Δ x) - \overset{0}{f (0)}}{Δ x} \\ = lim_{Δ x \to 0} \frac{(Δ x)^{2} \sin \frac{1}{Δ x}}{Δ x} \\ = lim_{Δ x \to 0} (Δ x \sin \frac{1}{Δ x}) \\ = 0. \end{aligned}$
[Recall that $lim_{h \to 0} h \sin \frac{1}{h} = 0$ . See the Section on Theorems for Calculating Limits for more information.]

When $x \neq 0$ , we can find $f^{'} (x)$ by using differentiation rules:
$f (x) = \underset{u}{\underset{⏟}{x^{2}}} \underset{v}{\underset{⏟}{\sin \frac{1}{x}}}$
$\begin{matrix} (i) & f^{'} (x) = \underset{u^{'}}{\underset{⏟}{2 x}} \underset{v}{\underset{⏟}{\sin \frac{1}{x}}} + \underset{u}{\underset{⏟}{x^{2}}} \underset{v^{'}}{\underset{⏟}{\frac{d}{d x} \sin \frac{1}{x}}} \end{matrix}$
To find $\frac{d}{d x} \sin \frac{1}{x}$ let $w = \frac{1}{x}$
$\begin{aligned} \frac{d}{d x} \sin \frac{1}{x} & = \frac{d}{d x} \sin w \\ = \frac{d}{d w} (\sin w) \times \frac{d w}{d x} \\ = (\cos w) (- \frac{1}{x^{2}}) \\ = (\cos \frac{1}{x}) (- \frac{1}{x^{2}}) \end{aligned}$
[ $\frac{d}{d x} \frac{1}{x} = \frac{d}{d x} x^{- 1} = - x^{- 2} = - \frac{1}{x^{2}}$ ]

Now we can simply plug the formula for $\frac{d}{d x} \sin \frac{1}{x}$ in (i)
$\begin{aligned} f^{'} (x) & = 2 x \sin \frac{1}{x} + x^{2} (- \frac{1}{x^{2}}) (\cos \frac{1}{x}) \\ = 2 x \sin \frac{1}{x} - \cos \frac{1}{x} (x \neq 0) \end{aligned}$
Therefore,
$f^{'} (x) = {\begin{cases} 2 x \sin \frac{1}{x} - \cos \frac{1}{x} & if x \neq 0 \\ 0 & if x = 0 \end{cases} .$
Because $\cos (1 / x)$ moves up and down so quickly as $x \to 0$ , it does not approach a number, and $lim_{x \to 0} \cos (1 / x)$ does not exists. Thus
$\begin{aligned} lim_{x \to 0} f^{'} (x) & = lim_{x \to 0} (2 x \sin \frac{1}{x} - \cos \frac{1}{x}) \\ = 2 lim_{x \to 0} x \sin \frac{1}{x} - lim_{x \to 0} \cos \frac{1}{x} \\ = 2 (0) - D N E \end{aligned}$
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 $x = 0$ , $f^{''} (0)$ does not exist.

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