7. Successive Differentiation

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):

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

    \((3) \dfrac{2}{\sqrt[3]{\theta^8}} – \dfrac{1.056}{\sqrt[5]{\theta^{11}}}\),     \(\dfrac{2.3232}{\sqrt[5]{\theta^{16}}} – \dfrac{16}{3 \sqrt[3]{\theta^{11}}}\).

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

    \((5) 12x + 2\),   \(12\).                                                 

    \((6) 6x^2 – 9x\),   \(12x – 9\).

    \((7)\) \(\begin{aligned}[t] &\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^5}}\right) +\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^7}} – \dfrac{1}{\sqrt{\theta^3}}\right). \\ &\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^5}} – \dfrac{1}{\sqrt{\theta^3}}\right) -\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^9}} + \dfrac{1}{\sqrt{\theta^7}}\right). \end{aligned}\)