Derivatives of Higher Order

Chapter Summary (Express)

Let us try the effect of repeating several times over the operation of differentiating a function (see the concept of a function). Begin with a concrete case.

Let $y = x^{5}$ . \begin{align} &\text{1st differentiation, } &&5x^4. && \\ &\text{2nd differentiation, } &&5 \times 4x^3 &&= 20x^3. \\ &\text{3rd differentiation, } &&5 \times 4 \times 3x^2 &&= 60x^2. \\ &\text{4th differentiation, } &&5 \times 4 \times 3 \times 2x &&= 120x. \\ &\text{5th differentiation, } &&5 \times 4 \times 3 \times 2 \times 1 &&= 120. \\ &\text{6th differentiation, } && &&= 0. \end{align}

There is a certain notation, with which we are already acquainted (see here), used by some writers, that is very convenient. This is to employ the general symbol $f (x)$ for any function of $x$ . Here the symbol $f ()$ is read as “function of,” without saying what particular function is meant. So the statement $y = f (x)$ merely tells us that $y$ is a function of $x$ , it may be $x^{2}$ or $a x^{n}$ , or $\cos x$ or any other complicated function of $x$ .

The corresponding symbol for the derivative is f^{\prime}(x), which is simpler to write than $\frac{d y}{d x}$ . This is called the derivative of the function $𝒇$ , the first derivative of $𝒇$ , or simply the derivative function. Instead of $\frac{d y}{d x}$ or f^\prime(x), we sometimes simply write y^\prime.

Suppose we differentiate over again, we shall get the second derivative of $𝒇$ (or $𝒚$ ) or the second order derivative of $𝒇$ (or $𝒚$ ), which is denoted by f^{\prime\prime}(x) or y^{\prime\prime}; and so on.

Now let us generalize.

Let $y = f (x) = x^{n}$ . \begin{align} &\text{1st differentiation,} &&y^\prime=f^\prime(x) = nx^{n-1}. \\ &\text{2nd differentiation,} &&y^{\prime\prime}=f^{\prime\prime}(x) = n(n-1)x^{n-2}. \\ &\text{3rd differentiation,} &&y^{\prime\prime\prime}=f^{\prime\prime\prime}(x) = n(n-1)(n-2)x^{n-3}. \\ &\text{4th differentiation,} &&y^{\prime\prime\prime\prime}=f^{\prime\prime\prime\prime}(x) = n(n-1)(n-2)(n-3)x^{n-4}. \\ &&\vdots \end{align}

In general, after differentiating the original function $n$ times, we obtain the $n$ -th derivative of $f$ or $y$ with respect to $x$ , which is also known as the derivative of order $n$ . When the differentiation order reaches four or more, rather than repeatedly using accents (also known as primes), a more streamlined approach is often adopted. The order of differentiation is denoted using parentheses, with the derivative order presented as a superscript to $f$ or $y$ . This notation is not only clearer but also helps reduce the risk of miscounting the number of primes. For instance, we often write $y^{(4)}$ or $f^{(4)} (x)$ instead of y^{\prime\prime\prime\prime} and f^{\prime\prime\prime\prime}(x).

There is another way of indicating successive differentiations. For, \begin{align} &\text{if the original function is } &&y = f(x); \\ &\text{once differentiating gives } &&\frac{dy}{dx} = f^{\prime}(x); \\ &\text{twice differentiating gives } &&\frac{d\left(\dfrac{dy}{dx}\right)}{dx} = f^{\prime\prime}(x); \end{align} and this is more conveniently written as $\frac{d^{2} y}{(d x)^{2}}$ , or more usually $\frac{d^{2} y}{d x^{2}}$ . Similarly, we may write as the result of differentiating three times, \dfrac{d^3y}{dx^3} = f^{\prime\prime\prime}(x).

How to Read the Symbols for Derivatives

f^\prime(x)	eff prime of eks
f^{\prime\prime} (x)	eff double prime of eks
f^{\prime\prime\prime}(x)	eff triple prime of eks
$f^{(n)} (x)$	eff super en of eks (or the en-th derivative of eff of eks)
y^\prime	wy prime
y^{\prime\prime}	wy double prime
y^{\prime\prime\prime}	wy triple prime
$y^{(n)}$	wy super en (or the en-th derivative of wy)
$\frac{d y}{d x}$	dee wy over dee eks
$\frac{d^{2} y}{d x^{2}}$	dee squared wy over dee eks squared

Examples

Now let us try $y = f (x) = 7 x^{4} + 3.5 x^{3} - \frac{1}{2} x^{2} + x - 2$ . \begin{align} \frac{dy}{dx} &= f^{\prime}(x) = 28x^3 + 10.5x^2 - x + 1, \\ \frac{d^2y}{dx^2} &= f^{\prime\prime}(x) = 84x^2 + 21x - 1, \\ \frac{d^3y}{dx^3} &= f^{\prime\prime\prime}(x) = 168x + 21, \\ \frac{d^4y}{dx^4} &= f^{(4)}(x) = 168, \\ \frac{d^5y}{dx^5} &= f^{(5)}(x) = 0. \end{align}

In a similar manner if $y = ϕ (x) = 3 x (x^{2} - 4)$ , \begin{align} \phi^\prime(x) &= \frac{dy}{dx} = 3\bigl[x \times 2x + (x^2 - 4) \times 1\bigr] = 3(3x^2 - 4), \\ \phi^{\prime\prime}(x) &= \frac{d^2y}{dx^2} = 3 \times 6x = 18x, \\ \phi^{\prime\prime\prime}(x) &= \frac{d^3y}{dx^3} = 18, \\ \phi^{(4)}(x) &= \frac{d^4y}{dx^4} = 0. \end{align}

Exercises

Find $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ for the following expressions:

Exercise 7.1. $y = 17 x + 12 x^{2}$ .

Answer

$17 + 24 x$ ; $24$ .

Solution

\begin{align} & y=17 x+12 x^{2} \\ & \frac{d y}{d x}=17+24 x \\ & \frac{d^{2} y}{d x^{2}}=24 \end{align}

Exercise 7.2. $y = \frac{x^{2} + a}{x + a}$ .

Answer

$\frac{x^{2} + 2 a x - a}{(x + a)^{2}}$ ; $\frac{2 a (a + 1)}{(x + a)^{3}}$ .

Solution

$y = \frac{x^{2} + a}{x + a}$ Using the Quotient Rule

\begin{align} \frac{d y}{d x} & =\frac{2 x(x+a)-\left(x^{2}+a\right)}{(x+a)^{2}} \\ & =\frac{x^{2}+2 a x-a}{(x+a)^{2}} \end{align}

To find $\frac{d^{2} y}{d x^{2}}$ , we use the Quotient Rule again.

\begin{align} \frac{d^{2} y}{d x^{2}}&=\frac{(2 x+2 a)(x+a)^{2}-\frac{d\left(x^{2}+2 a x+a\right)^{2}}{d x}\left(x^{2}+2 a x-a\right)}{(x+a)^{4}}\\ & =\frac{2(x+a)^{3}-(2 x+2 a)\left(x^{2}+2 a x-a\right)}{(x+a)^{4}} \\ & =\frac{2(x+a)\left[(x+a)^{2}-\left(x^{2}+2 a x-a\right)\right]}{(x+a)^{4}}\\ & =\frac{2\left[x^{2}+2 a x+a^{2}-x^{2}-2 a x+a\right]}{(x+a)^{3}} \\ & =\frac{2\left(a^{2}+a\right)}{(x+a)^{3}} \\ & =\frac{2 a(a+1)}{(x+a)^{3}} \end{align}

Exercise 7.3. $y = 1 + \frac{x}{1} + \frac{x^{2}}{1 \times 2} + \frac{x^{3}}{1 \times 2 \times 3} + \frac{x^{4}}{1 \times 2 \times 3 \times 4}$ .

Answer

$1 + x + \frac{x^{2}}{1 \times 2} + \frac{x^{3}}{1 \times 2 \times 3}$ ; $1 + x + \frac{x^{2}}{1 \times 2}$ .

Solution

\begin{align} y & =1+\frac{x}{1}+\frac{x^{2}}{1 \times 2}+\frac{x^{3}}{1 \times 2 \times 3}+\frac{x^{4}}{1 \times 2 \times 3 \times 4}. \\ \frac{d y}{d x} & =1+\frac{x}{1}+\frac{x^{2}}{1 \times 2}+\frac{x^{3}}{1 \times 2 \times 3}. \\ \frac{d^{2} y}{d x} & =1+\frac{x}{1}+\frac{x^{2}}{1 \times 2}. \end{align}

Exercise 7.4. Find the 2nd and 3rd derivatives in the Exercises of Chapter 6, No. 1 to No. 7:

Expressions:

First Exercise:
1. $u = 1 + x + \frac{x^{2}}{1 \times 2} + \frac{x^{3}}{1 \times 2 \times 3} + \dots$ .
2. $y = a x^{2} + b x + c$ .
3. $y = (x + a)^{2}$ .
4. $y = (x + a)^{3}$ .
$w = a t - \frac{1}{2} b t^{2}$ .
$y = (x + \sqrt{- 1}) \times (x - \sqrt{- 1}) .$
$y = (197 x - 34 x^{2}) \times (7 + 22 x - 83 x^{3}) .$
$x = (y + 3) \times (y + 5)$ .
$y = 1.3709 x \times (112.6 + 45.202 x^{2})$ .
$y = \frac{2 x + 3}{3 x + 2}$ .

and in Example 6.4 to Example 6.10:

Example 6.4: $y = \frac{a}{b^{2}} x^{3} - \frac{a^{2}}{b} x + \frac{a^{2}}{b^{2}}$ .

Example 6.5: $y = 2 a \sqrt{b x^{3}} - \frac{3 b \sqrt[3]{a}}{x} - 2 \sqrt{a b}$

Example 6.6: $z = 1.8 \sqrt[3]{\frac{1}{θ^{2}}} - \frac{4.4}{\sqrt[5]{θ}} - 27$ .

Example 6.7: $v = (3 t^{2} - 1.2 t + 1)^{3}$

Example 6.8: $y = (2 x - 3) (x + 1)^{2}$ .

Example 6.9: $y = 0.5 x^{3} (x - 3)$ .

Example 6.10: $w = (θ + \frac{1}{θ}) (\sqrt{θ} + \frac{1}{\sqrt{θ}})$ .

Answer

(Exercises of Chapter 6):

(1) $\frac{d^{2} y}{d x^{2}} = \frac{d^{3} y}{d x^{3}} = 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots$ .
(2) $2 a$ , $0$ .
(3) $2$ , $0$ .
(4) $6 x + 6 a$ , $6$ .

$- b$ , $0$ .

$2$ , $0$ .

$\begin{matrix} 56440 x^{3} - 196212 x^{2} - 4488 x + 8192. \\ 169320 x^{2} - 392424 x - 4488. \end{matrix}$

$2$ , $0$ .

$371.80453 x$ , $371.80453$ .

$\frac{30}{(3 x + 2)^{3}}$ , $- \frac{270}{(3 x + 2)^{4}}$ .

Example 6.4: $\frac{6 a}{b^{2}} x$ , $\frac{6 a}{b^{2}}$ .

Example 6.5: $\frac{3 a \sqrt{b}}{2 \sqrt{x}} - \frac{6 b \sqrt[3]{a}}{x^{3}}$ , $\frac{18 b \sqrt[3]{a}}{x^{4}} - \frac{3 a \sqrt{b}}{4 \sqrt{x^{3}}}$ .

Example 6.6: $\frac{2}{\sqrt[3]{θ^{8}}} - \frac{1.056}{\sqrt[5]{θ^{11}}}$ , $\frac{2.3232}{\sqrt[5]{θ^{16}}} - \frac{16}{3 \sqrt[3]{θ^{11}}}$ .

Example 6.7: $810 t^{4} - 648 t^{3} + 479.52 t^{2} - 139.968 t + 26.64 .$ , $3240 t^{3} - 1944 t^{2} + 959.04 t - 139.968 .$

Example 6.8: $12 x + 2$ , $12$ .

Example 6.9: $6 x^{2} - 9 x$ , $12 x - 9$ .

Example 6.10: \begin{align} &\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^5}}\right) +\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^7}} - \dfrac{1}{\sqrt{\theta^3}}\right). \\ &\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^5}} - \dfrac{1}{\sqrt{\theta^3}}\right) -\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^9}} + \dfrac{1}{\sqrt{\theta^7}}\right). \end{align}

Solution

(1)

(a) We learned that

$\frac{d u}{d x} = u$

Therefore,

$\frac{d^{2} u}{d x^{2}} = \frac{d (\frac{d u}{d x})}{d x} = \frac{d u}{d x} = u$

and

$\frac{d^{3} u}{d x^{3}} = \frac{d (\frac{d^{2} u}{d x^{2}})}{d x} = \frac{d u}{d x} = u .$

(b) Since $\frac{d y}{d x} = 2 a x + b$ then \begin{align} & \frac{d^{2} y}{d x^{2}}=2 a \\ & \frac{d^{3} y}{d x^{3}}=0 \end{align}

(c) Since $\frac{d y}{d x} = 2 (x + a) = 2 x + 2 a$

\begin{align} & \frac{d^{2} y}{d x^{2}}=2 \\ & \frac{d^{3} y}{d x^{3}}=0 \end{align}

(d) Since

\begin{align} & \frac{d y}{d x}=3(x+a)^{2}=3\left(x^{2}+2 a x+a^{2}\right) \\ & \frac{d^{2} y}{d x^{2}}=6 x+6 a=6(x+a) \\ & \frac{d^{3} y}{d x^{3}}=6 \end{align}

(2) Since $\frac{d w}{d t} = a - b t$

\begin{align} & \frac{d^{2} w}{d t^{2}}=-b \\ & \frac{d^{3} w}{d t^{3}}=0 \end{align}

(3) Since $\frac{d y}{d x} = 2 x,$ \begin{align} & \frac{d^{2} y}{d x^{2}}=2 \quad \text { and } \quad \frac{d^{3} y}{d x^{3}}=0 \end{align}

(4) Since $\frac{d y}{d x} = 14110 x^{4} - 65404 x^{3} - 22404 x^{2} + 8192 x + 1379$ ,

\begin{align} \frac{d^{2} y}{d x^{2}} & =56440 x^{3}-196212 x^{2}-44808 x+8192 \\ \frac{d^{3} y}{d x^{3}} & =3 \times 56440 x^{2}-2 \times 196212 x-44808 \\ & =169320 x^{2}-392424 x-44808 \end{align}

(5) Since $\frac{d x}{d y} = 2 y + 5$

$\frac{d^{2} x}{d y^{2}} = 2 and \frac{d^{3} x}{d y^{3}} = 0$

(6) Since $\frac{d y}{d x} = 185.9022654 x^{2} + 154.36334$

\begin{align} & \frac{d^2 y}{d x^2}=2 \times 185.9022654 x=371.8045308 x \\ & \frac{d^3 y}{d x^3}=371.8045308 \end{align}

(7) Since $\frac{d y}{d x} = - \frac{5}{(3 x + 2)^{2}}$ , \begin{align} \frac{d^{2} y}{d x^{2}} & =-\frac{0 \times(3 x+2)^{2}-5 \frac{d\left[(3 x+2)^{2}\right]}{d x}}{(3 x+2)^{4}} \\ & =\frac{5 \frac{d\left[9 x^{2}+12 x+4\right]}{d x}}{(3 x+2)^{4}} \\ & =\frac{5(18 x+12)}{(3 x+2)^{4}} \\ & =\frac{30(3 x+2)}{(3 x+2)^{4}} \\ & =\frac{30}{(3 x+2)^{3}} \\ \frac{d^{3} y}{d x^{3}} & =\frac{-30 \frac{d\left[(3 x+2)^{3}\right]}{d x}}{(3 x+2)^{6}} \\ & =-\frac{30 \frac{d\left(27 x^{3}+54 x^{2}+36 x+8\right)}{d x}}{(3 x+2)^{6}} \\ & =\frac{30\left(81 x^{2}+108 x+36\right)}{(3 x+2)^{6}} \end{align}

\begin{align} & =\frac{30 \times 9\left(9 x^{2}+12 x+4\right)}{(3 x+2)^{6}} \\ & =\frac{270(3 x+2)^{2}}{(3 x+2)^{6}} \\ & =\frac{270}{(3 x+2)^{5}} \end{align}

Example 19. Since $\frac{d y}{d x} = \frac{3 a}{b^{2}} x^{2} - \frac{a^{2}}{b}$

$\frac{d^{2} y}{d x^{2}} = \frac{6 a}{b^{2}} x and \frac{d^{3} y}{d x^{3}} = \frac{6 a}{b^{2}}$

Example 20. Since $\frac{d y}{d x} = 3 a \sqrt{b x} + \frac{3 b \sqrt[3]{a}}{x^{2}}$ . We may rewrite it as $\frac{d y}{d x} = 3 a \sqrt{b} x^{\frac{1}{2}} + 3 b \sqrt[3]{a} x^{- 2}$ Therefore \begin{align} \frac{d^2 y}{d x^2} & =\frac{1}{2} 3 a \sqrt{b} x^{-\frac{1}{2}}-6 b \sqrt[3]{a} x^{-3} \\ & =\frac{3}{2} a \sqrt{\frac{b}{x}}-\frac{6 b \sqrt[3]{a}}{x^3} \end{align} and

\begin{align} \frac{d^{3} y}{d x^{3}} & =-\frac{1}{4} 3 a \sqrt{b} x^{-\frac{3}{2}}+18 b \sqrt[3]{a} x^{-4} \\ & =-\frac{3 a \sqrt{b}}{4 \sqrt{x^{3}}}+\frac{18 b \sqrt[4]{a}}{x^{4}} \end{align}

Example 21. Since $\frac{d z}{d θ} = - 1.2 θ^{- \frac{5}{3}} + 0.88 θ^{- \frac{6}{5}}$

\begin{align} \frac{d^{2} z}{d \theta^{2}} & =2 \theta^{-\frac{8}{3}}-1.056 \theta^{-\frac{11}{5}} \\ & =\frac{2}{\sqrt[3]{\theta^{8}}}-\frac{1.056}{\sqrt[5]{\theta^{11}}} \\ \frac{d^{3} z}{d \theta^{3}} & =-\frac{16}{3} \theta^{-\frac{11}{3}}+2.3232 \theta^{-\frac{16}{5}} \\ & =-\frac{16}{\sqrt[3]{\theta^{11}}}+\frac{2.3232}{\sqrt[5]{\theta^{16}}} \end{align}

Example 22. Since $\frac{d v}{d t} = 162 t^{5} - 162 t^{4} + 159.84 t^{3} - 69.984 t^{2} + 26.64 t$

\begin{align} & \frac{d^{2} v}{d t^{2}}=810 t^{4}-648 t^{3}+479.52 t^{2}-139.968 t+26.64 \\ & \frac{d^{3} v}{d t^{3}}=3240 t^{3}-1944 t^{2}+959.04 t-139.968 \end{align}

Example 23. Since $\frac{d y}{d x} = 2 (x + 1) (3 x - 2)$

$\frac{d^{2} y}{d x^{2}} = 2 (3 x - 2) + 6 (x + 1) = 12 x + 2$

$\frac{d^{3} y}{d x^{3}} = 12$

Example 24. Since $\frac{d y}{d x} = 2 x^{3} - 4.5 x^{2}$

\begin{align} & \frac{d^{2} y}{d x^{2}}=6 x^{2}-9 x \\ & \frac{d^{3} y}{d x^{3}}=12 x-9 \end{align}

Example 25. Since $\frac{d ω}{d θ} = \frac{3}{2} (\sqrt{θ} - \frac{1}{\sqrt{θ^{5}}}) + \frac{1}{2} (\frac{1}{\sqrt{θ}} - \frac{1}{\sqrt{θ^{3}}})$ , we can rewrite it as

$\frac{d w}{d θ} = \frac{3}{2} (θ^{\frac{1}{2}} - θ^{- \frac{5}{2}}) + \frac{1}{2} (θ^{- \frac{1}{2}} - θ^{- \frac{3}{2}})$ Therefore

\begin{align} \frac{d^{2} w}{d \theta^{2}} & =\frac{3}{2}\left(\frac{1}{2} \theta^{-\frac{1}{2}}+\frac{5}{2} \theta^{-\frac{7}{2}}\right)+\frac{1}{2}\left(-\frac{1}{2} \theta^{-\frac{3}{2}}+\frac{3}{2} \theta^{-\frac{5}{2}}\right) \\ & =\frac{3}{4} \theta^{-\frac{1}{2}}+\frac{15}{4} \theta^{-\frac{7}{2}}-\frac{1}{4} \theta^{-\frac{3}{2}}+\frac{3}{4} \theta^{-\frac{5}{2}} \\ & =\frac{3}{4}\left(\theta^{-\frac{1}{2}}+\theta^{-\frac{5}{2}}\right)+\frac{1}{4}\left(15 \theta^{-\frac{7}{2}}-\theta^{-\frac{3}{2}}\right) \\ & =\frac{3}{4}\left(\frac{1}{\sqrt{\theta}}+\frac{1}{\sqrt{\theta^{5}}}\right)+\frac{1}{4}\left(\frac{15}{\sqrt{\theta^{7}}}-\frac{1}{\sqrt{\theta^{3}}}\right) \end{align}

\begin{align} \frac{d^{3} w}{d \theta^{3}} & =\frac{3}{4}\left(-\frac{1}{2} \theta^{\frac{2}{2}}-\frac{5}{2} \theta^{2}\right)+\frac{1}{4}\left(-\frac{105}{2} \theta^{-\frac{2}{2}}+\frac{3}{2} \theta^{-\frac{1}{2}}\right) \\ & =\frac{3}{8}\left(\theta^{-\frac{5}{2}}-\theta^{-\frac{3}{2}}\right)-\frac{15}{8}\left(7 \theta^{-\frac{9}{2}}+\theta^{-\frac{7}{2}}\right) \\ & =\frac{3}{8}\left(\frac{1}{\sqrt{\theta^{5}}}-\frac{1}{\sqrt{\theta^{3}}}\right)-\frac{15}{8}\left(\frac{7}{\sqrt{\theta^{9}}}+\frac{1}{\sqrt{\theta^{7}}}\right) \end{align}

Full Chapter

Let us try the effect of repeating several times over the operation of differentiating a function (see the concept of a function). Begin with a concrete case.

Let $y = x^{5}$ . \begin{align} &\text{First differentiation, } &&5x^4. && \\ &\text{Second differentiation, } &&5 \times 4x^3 &&= 20x^3. \\ &\text{Third differentiation, } &&5 \times 4 \times 3x^2 &&= 60x^2. \\ &\text{Fourth differentiation, } &&5 \times 4 \times 3 \times 2x &&= 120x. \\ &\text{Fifth differentiation, } &&5 \times 4 \times 3 \times 2 \times 1 &&= 120. \\ &\text{Sixth differentiation, } && &&= 0. \end{align}

There is a certain notation, with which we are already acquainted, used by some writers, that is very convenient. This is to employ the general symbol $f (x)$ for any function of $x$ . Here the symbol $f ()$ is read as “function of,” without saying what particular function is meant. So the statement $y = f (x)$ merely tells us that $y$ is a function of $x$ , it may be $x^{2}$ or $a x^{n}$ , or $\cos x$ or any other complicated function of $x$ .

The corresponding symbol for the derivative is f^{\prime}(x), which is simpler to write than $\frac{d y}{d x}$ . This is called the derivative of $𝒚$ with respect to $𝒙$ , the derivative of the function $𝒇$ , or simply the derivative function. Instead of $\frac{d y}{d x}$ or f^\prime(x), we sometimes simply write y^\prime.

Suppose we differentiate over again, we shall get the second derivative of $𝒇$ or the second derivative of $𝒚$ with respect to $𝒙$ , which is denoted by f^{\prime\prime}(x) or y^{\prime\prime}; and so on.

Now let us generalize.

Let $y = f (x) = x^{n}$ . \begin{align} &\text{First differentiation,} &&y^\prime=f^\prime(x) = nx^{n-1}. \\ &\text{Second differentiation,} &&y^{\prime\prime}=f^{\prime\prime}(x) = n(n-1)x^{n-2}. \\ &\text{Third differentiation,} &&y^{\prime\prime\prime}=f^{\prime\prime\prime}(x) = n(n-1)(n-2)x^{n-3}. \\ &\text{Fourth differentiation,} &&y^{\prime\prime\prime\prime}=f^{\prime\prime\prime\prime}(x) = n(n-1)(n-2)(n-3)x^{n-4}. \\ &&\vdots \end{align}

In general, after differentiating the original function $n$ times, we obtain the $n$ -th derivative of $f$ or $y$ with respect to $x$ , also known as the derivative of order $n$ . When the differentiation order reaches four or more, rather than repeatedly using accents (also known as primes), a more streamlined approach is often adopted. The order of differentiation is denoted using parentheses, with the derivative order presented as a superscript to $f$ or $y$ . This notation is not only clearer but also helps reduce the risk of miscounting the number of primes. For instance, we often write $y^{(4)}$ or $f^{(4)} (x)$ instead of y^{\prime\prime\prime\prime} and f^{\prime\prime\prime\prime}(x).

How to Read the Symbols for Derivatives

f^\prime(x)	eff prime of eks
f^{\prime\prime} (x)	eff double prime of eks
f^{\prime\prime\prime}(x)	eff triple prime of eks
$f^{(n)} (x)$	eff super en of eks (or the en-th derivative of eff of eks)
y^\prime	wy prime
y^{\prime\prime}	wy double prime
y^{\prime\prime\prime}	wy triple prime
$y^{(n)}$	why super en (or the en-th derivative of wy)
$\frac{d y}{d x}$	dee wy over dee eks
$\frac{d^{2} y}{d x^{2}}$	dee squared wy over dee eks squared

Examples

Exercises

Find $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ for the following expressions:

Exercise 7.1. $y = 17 x + 12 x^{2}$ .

Answer

$17 + 24 x$ ; $24$ .

Solution

\begin{align} & y=17 x+12 x^{2} \\ & \frac{d y}{d x}=17+24 x \\ & \frac{d^{2} y}{d x^{2}}=24 \end{align}

Exercise 7.2. $y = \frac{x^{2} + a}{x + a}$ .

Answer

$\frac{x^{2} + 2 a x - a}{(x + a)^{2}}$ ; $\frac{2 a (a + 1)}{(x + a)^{3}}$ .

Solution

$y = \frac{x^{2} + a}{x + a}$ Using the Quotient Rule

\begin{align} \frac{d y}{d x} & =\frac{2 x(x+a)-\left(x^{2}+a\right)}{(x+a)^{2}} \\ & =\frac{x^{2}+2 a x-a}{(x+a)^{2}} \end{align}

To find $\frac{d^{2} y}{d x^{2}}$ , we use the Quotient Rule again.

Exercise 7.3. $y = 1 + \frac{x}{1} + \frac{x^{2}}{1 \times 2} + \frac{x^{3}}{1 \times 2 \times 3} + \frac{x^{4}}{1 \times 2 \times 3 \times 4}$ .

Answer

$1 + x + \frac{x^{2}}{1 \times 2} + \frac{x^{3}}{1 \times 2 \times 3}$ ; $1 + x + \frac{x^{2}}{1 \times 2}$ .

Solution

Exercise 7.4. Find the 2nd and 3rd derivatives in the Exercises of Chapter 6, No. 1 to No. 7:

Expressions:

First Exercise:
1. $u = 1 + x + \frac{x^{2}}{1 \times 2} + \frac{x^{3}}{1 \times 2 \times 3} + \dots$ .
2. $y = a x^{2} + b x + c$ .
3. $y = (x + a)^{2}$ .
4. $y = (x + a)^{3}$ .
$w = a t - \frac{1}{2} b t^{2}$ .
$y = (x + \sqrt{- 1}) \times (x - \sqrt{- 1}) .$
$y = (197 x - 34 x^{2}) \times (7 + 22 x - 83 x^{3}) .$
$x = (y + 3) \times (y + 5)$ .
$y = 1.3709 x \times (112.6 + 45.202 x^{2})$ .
$y = \frac{2 x + 3}{3 x + 2}$ .