Listen to this text
 Listen to this text
CHAPTER VII.
SUCCESSIVE DIFFERENTIATION.

Let us try the effect of repeating several times over the operation of differentiating a function (see p. 14). Begin with a concrete case.

Let $y=x^{5}$ .

 First differentiation, $5x^{4}$ . Second differentiation, $5\times 4x^{3}$ $=20x^{3}$ . Third differentiation, $5\times 4\times 3x^{2}$ $=60x^{2}$ . Fourth differentiation, $5\times 4\times 3\times 2x$ $=120x$ . Fifth differentiation, $5\times 4\times 3\times 2\times 1$ $=120$ . Sixth differentiation, $=0$ .

There is a certain notation, with which we are already acquainted (see p. 15), 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 ${\frac {dy}{dx}}$ . 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 $y=f(x)=x^{n}$ .

 First differentiation, $f'(x)=nx^{n-1}$ . Second differentiation, $f''(x)=n(n-1)x^{n-2}$ . Third differentiation, $f'''(x)=n(n-1)(n-2)x^{n-3}$ . Fourth differentiation, $f''''(x)=n(n-1)(n-2)(n-3)x^{n-4}$ . etc., etc.

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

if the original function be $y=f(x)$ ;

once differentiating gives ${\frac {dy}{dx}}=f'(x)$ ;

twice differentiating gives ${\frac {d({\frac {dy}{dx}})}{dx}}=f''(x)$ ;

and this is more conveniently written as ${\frac {d^{2}y}{(dx)^{2}}}$ , or more usually ${\frac {d^{2}y}{dx^{2}}}$ . Similarly, we may write as the result of thrice differentiating, ${\frac {d^{3}y}{dx^{3}}}=f'''(x)$ .

Examples.

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

${\frac {dy}{dx}}=f'(x)=28x^{3}+10.5x^{2}-x+1$ ,
${\frac {d^{2}y}{dx^{2}}}=f''(x)=84x^{2}+21x-1$ ,
${\frac {d^{3}y}{dx^{3}}}=f'''(x)=168x+21$ ,
${\frac {d^{4}y}{dx^{4}}}=f''''(x)=168$ ,
${\frac {d^{5}y}{dx^{5}}}=f'''''(x)=0$ .

In a similar manner if $y=\phi (x)=3x(x^{2}-4)$ ,

$\phi '(x)={\frac {dy}{dx}}=3\left[x\times 2x+(x^{2}-4)\times 1\right]=3(3x^{2}-4)$ ,
$\phi ''(x)={\frac {d^{2}y}{dx^{2}}}=3\times 6x=18x$ ,
$\phi '''(x)={\frac {d^{3}y}{dx^{3}}}=18$ ,
$\phi ''''(x)={\frac {d^{4}y}{dx^{4}}}=0$ .

Exercises IV. (See page 255 for Answers.) Find ${\frac {dy}{dx}}$ and ${\frac {d^{2}y}{dx^{2}}}$ for the following expressions:

(1) $y=17x+12x^{2}$ .
(2) $y={\frac {x^{2}+a}{x+a}}$ .
(3) $y=1+{\frac {x}{1}}+{\frac {x^{2}}{1\times 2}}+{\frac {x^{3}}{1\times 2\times 3}}+{\frac {x^{4}}{1\times 2\times 3\times 4}}$ .
(4) Find the 2nd and 3rd derived functions in the Exercises III. (p. 46), No. 1 to No. 7, and in the Examples given (p. 41), No. 1 to No. 7.