Calculus Made Easy/Chapter 7

by Silvanus Phillips Thompson

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

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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 $y=\phi (x)=3x(x^{2}-4)$,

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

$\phi ''(x)={\frac {d^{2}y}{dx^{2}}}=3\times 6x=18x$,

$\phi '''(x)={\frac {d^{3}y}{dx^{3}}}=18$,

$\phi ''''(x)={\frac {d^{4}y}{dx^{4}}}=0$.

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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.

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