Let us try the effect of repeating several times over the operation of differentiating a function (see chapter 3). Begin with a concrete case.

Let $$y = x^5$$. \begin{aligned} &\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{aligned}

There is a certain notation, with which we are already acquainted (see chapter 3), 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 $$f^{\prime\prime}(x)$$; and so on.

Now let us generalize.

Let $y = f(x) = x^n$.

\begin{aligned} &\text{First differentiation, } &&f'(x)=nx^{n-1}. \\ &\text{Second differentiation, } &&f^{\prime\prime}(x)=n(n-1)x^{n-2}. \\ &\text{Third differentiation, }&& f^{\prime\prime\prime}(x)=5 n(n-1)(n-2)x^{n-3}. \\ &\text{Fourth differentiation, } &&f^{\prime\prime\prime\prime}(x)=n(n-1)(n-2)(n-3)x^{n-4}.\\ & &&etc.\end{aligned}

But this is not the only way of indicating successive differentiations. For,

\begin{aligned} &\text{if the original function be} &&y=f(x) \\ &\text{once differentiating gives} && \dfrac{dy}{dx}= f'(x) \\ &\text{twice differentiating gives} && \dfrac{d\dfrac{dy}{dx}}{dx}= f^{\prime\prime}(x)\end{aligned}

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, $$\dfrac{d^3y}{dx^3} = f^{\prime\prime\prime}(x)$$.

Examples.

Example 1
Now let us try $$y = f(x) = 7x^4 + 3.5x^3 – \frac{1}{2}x^2 + x – 2$$.
Solution

\begin{aligned} \frac{dy}{dx} &= f'(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^{\prime\prime\prime\prime}(x) = 168, \\ \frac{d^5y}{dx^5} &= f^{\prime\prime\prime\prime\prime}(x) = 0.\end{aligned}

Example 2
In a similar manner if $$y = \phi(x) = 3x(x^2 – 4)$$,
Solution

\begin{aligned} \phi'(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^{\prime\prime\prime\prime}(x) &= \frac{d^4y}{dx^4} = 0.\end{aligned}

Exercises IV.

Find $$\dfrac{dy}{dx}$$ and $$\dfrac{d^2y}{dx^2}$$ for the following expressions:

 $$(1) y = 17x + 12x^2$$ $$(2) y = \dfrac{x^2 + a}{x + a}$$

$$(3) y = 1 + \dfrac{x}{1} + \dfrac{x^2}{1\times2} + \dfrac{x^3}{1\times2\times3} + \dfrac{x^4}{1\times2\times3\times4}$$.

$$(4)$$ Find the 2nd and 3rd derived functions in the Exercises (chapter 6), No. 1 to No. 7, and in the Examples given (chapter 6), No. 1 to No. 7.

Solution
 $$(1) 17 + 24x$$;         $$24$$. $$(2) \dfrac{x^2 + 2ax – a}{(x + a)^2}$$;         $$\dfrac{2a(a + 1)}{(x + a)^3}$$

$$(3) 1 + x + \dfrac{x^2}{1 \times 2} + \dfrac{x^3}{\times 2 \times 3}$$;           $$1 + x + \dfrac{x^2}{1 \times 2}$$.

$$(4)$$ (Exercises, chapter 6):

• $$(1)$$ (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) $$6x + 6a$$, $$6$$.

• $$(2)$$ $$-b$$, $$0$$.                                         $$(3)$$ $$2$$, $$0$$.

• $$(4)$$ $$\begin{gathered}[t] 56440x^3 – 196212x^2 – 4488x + 8192. \\ 169320x^2 – 392424x – 4488. \end{gathered}$$

• $$(6) 2$$,   $$0$$.                                             $$(6)$$ $$371.80453x$$,   $$371.80453$$.

• $$(7) \dfrac{30}{(3x + 2)^3}$$,   $$-\dfrac{270}{(3x + 2)^4}$$.

(Examples, chapter 6):

• $$(1) \dfrac{6a}{b^2} x$$,    $$\dfrac{6a}{b^2}$$.                   $$(2) \dfrac{3a \sqrt{b}} {2 \sqrt{x}} – \dfrac{6b \sqrt{a}}{x^3}$$,     $$\dfrac{18b \sqrt{a}}{x^4} – \dfrac{3a \sqrt{b}}{4 \sqrt{x^3}}$$

$$(3) \dfrac{2}{\sqrt{\theta^8}} – \dfrac{1.056}{\sqrt{\theta^{11}}}$$,     $$\dfrac{2.3232}{\sqrt{\theta^{16}}} – \dfrac{16}{3 \sqrt{\theta^{11}}}$$.

$$(4)$$ $$\begin{gathered}[t] 810t^4 – 648t^3 + 479.52t^2 – 139.968t + 26.64. \\ 3240t^3 – 1944t^2 + 959.04t – 139.968. \end{gathered}$$

$$(5) 12x + 2$$,   $$12$$.

$$(6) 6x^2 – 9x$$,   $$12x – 9$$.

$$(7)$$ \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}