Let us try the effect of repeating several times over the operation of differentiating a function (see here). Begin with a concrete case.
Let $y = x^5$. \begin{alignat*}{3} &\text{First differentiation, } &&5x^4. && \\ &\text{Second differentiation, } &&5 × 4x^3 &&= 20x^3. \\ &\text{Third differentiation, } &&5 × 4 × 3x^2 &&= 60x^2. \\ &\text{Fourth differentiation, } &&5 × 4 × 3 × 2x &&= 120x. \\ &\text{Fifth differentiation, } &&5 × 4 × 3 × 2 × 1 &&= 120. \\ &\text{Sixth differentiation, } && &&= 0. \end{alignat*}
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 $ax^n$, or $\cos x$ or any other complicated function of $x$.
The corresponding symbol for the differential coefficient is $f'(x)$, which is simpler to write than $\dfrac{dy}{dx}$. This is called the “derived function” of $x$.
Suppose we differentiate over again, we shall get the “second derived function” or second differential coefficient, which is denoted by $f''(x)$; and so on.
Now let us generalize.
Let $y = f(x) = x^n$. \begin{align*} \text{First differentiation,}\; f'(x) &= nx^{n-1}. \\ \text{Second differentiation,}\; f''(x) &= n(n-1)x^{n-2}. \\ \text{Third differentiation,}\; f'''(x) &= n(n-1)(n-2)x^{n-3}. \\ \text{Fourth differentiation,}\; f''''(x) &= n(n-1)(n-2)(n-3)x^{n-4}. \\ &\llap{\text{etc.,}} \text{ etc.} \end{align*}
But this is not the only way of indicating successive differentiations. For, \begin{align*} \text{if the original function be}\; y &= f(x); \\ \text{once differentiating gives}\; \frac{dy}{dx} &= f'(x); \\ \text{twice differentiating gives}\; \frac{d\left(\dfrac{dy}{dx}\right)}{dx} &= f''(x); \end{align*} and this is more conveniently written as $\dfrac{d^2y}{(dx)^2}$, or more usually $\dfrac{d^2y}{dx^2}$. Similarly, we may write as the result of thrice differentiating, $\dfrac{d^3y}{dx^3} = f'''(x)$.
Examples Now let us try $y = f(x) = 7x^4 + 3.5x^3 - \frac{1}{2}x^2 + x - 2$. \begin{align*} \frac{dy}{dx} &= f'(x) = 28x^3 + 10.5x^2 - x + 1, \\ \frac{d^2y}{dx^2} &= f''(x) = 84x^2 + 21x - 1, \\ \frac{d^3y}{dx^3} &= f'''(x) = 168x + 21, \\ \frac{d^4y}{dx^4} &= f''''(x) = 168, \\ \frac{d^5y}{dx^5} &= f'''''(x) = 0. \end{align*} In a similar manner if $y = \phi(x) = 3x(x^2 - 4)$, \begin{align*} \phi'(x) &= \frac{dy}{dx} = 3\bigl[x × 2x + (x^2 - 4) × 1\bigr] = 3(3x^2 - 4), \\ \phi''(x) &= \frac{d^2y}{dx^2} = 3 × 6x = 18x, \\ \phi'''(x) &= \frac{d^3y}{dx^3} = 18, \\ \phi''''(x) &= \frac{d^4y}{dx^4} = 0. \end{align*}
Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for the following expressions:
(1) $y = 17x + 12x^2$.
(2) $y = \dfrac{x^2 + a}{x + a}$.
(3) $y = 1 + \dfrac{x}{1} + \dfrac{x^2}{1×2} + \dfrac{x^3}{1×2×3} + \dfrac{x^4}{1×2×3×4}$.
(4) Find the 2nd and 3rd derived functions in the Exercises III. (here), No. 1 to No. 7, and in the Examples given (here), No. 1 to No. 7.
(1) $17 + 24x$; $24$.
(2) $\dfrac{x^2 + 2ax - a}{(x + a)^2}$; $\dfrac{2a(a + 1)}{(x + a)^3}$.
(3) $1 + x + \dfrac{x^2}{1 × 2} + \dfrac{x^3}{1 × 2 × 3}$; $1 + x + \dfrac{x^2}{1 × 2}$.
Exercises III
(4) (Exercises III. ):
(1) (a ) $\dfrac{d^2 y}{dx^2} = \dfrac{d^3 y}{dx^3} = 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \ldots$.
(b ) $2a$, $0$.
(c ) $2$, $0$.
(d ) $6x + 6a$, $6$.
(2) $-b$, $0$.
(3) $2$, $0$.
(4) $\begin{gathered}[t] 56440x^3 - 196212x^2 - 4488x + 8192. \\ 169320x^2 - 392424x - 4488. \end{gathered}$
(5) $2$, $0$.
(6) $371.80453x$, $371.80453$.
(7) $\dfrac{30}{(3x + 2)^3}$, $-\dfrac{270}{(3x + 2)^4}$.
Examples ):
(1) $\dfrac{6a}{b^2} x$, $\dfrac{6a}{b^2}$.
(2) $\dfrac{3a \sqrt{b}} {2 \sqrt{x}} - \dfrac{6b \sqrt[3]{a}}{x^3}$, $\dfrac{18b \sqrt[3]{a}}{x^4} - \dfrac{3a \sqrt{b}}{4 \sqrt{x^3}}$.
(3) $\dfrac{2}{\sqrt[3]{\theta^8}} - \dfrac{1.056}{\sqrt[5]{\theta^{11}}}$, $\dfrac{2.3232}{\sqrt[5]{\theta^{16}}} - \dfrac{16}{3 \sqrt[3]{\theta^{11}}}$.
(4) $\begin{gathered}[t] 810t^4 - 648t^3 + 479.52t^2 - 139.968t + 26.64. \\ 3240t^3 - 1944t^2 + 959.04t - 139.968. \end{gathered}$
(5) $12x + 2$, $12$.
(6) $6x^2 - 9x$, $12x - 9$.
(7)
$\begin{aligned}[t]
&\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^5}}\right)
+\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^7}} - \dfrac{1}{\sqrt{\theta^3}}\right). \\
&\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^5}} - \dfrac{1}{\sqrt{\theta^3}}\right)
-\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^9}} + \dfrac{1}{\sqrt{\theta^7}}\right).
\end{aligned}$