Calculus Made Easy/Chapter 7

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 ${\displaystyle y=x^{5}}$.

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

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 ${\displaystyle f(x)}$ for any function of ${\displaystyle x}$. Here the symbol ${\displaystyle f(\;\;)}$ is read as “function of,” without saying what particular function is meant. So the statement ${\displaystyle y=f(x)}$ merely tells us that ${\displaystyle y}$ is a function of ${\displaystyle x}$, it may be ${\displaystyle x^{2}}$ or ${\displaystyle ax^{n}}$, or ${\displaystyle \cos {x}}$ or any other complicated function of ${\displaystyle x}$.

The corresponding symbol for the differential coefficient is ${\displaystyle f'(x)}$, which is simpler to write than ${\displaystyle {\frac {dy}{dx}}}$. This is called the “derived function” of ${\displaystyle x}$.

​Suppose we differentiate over again, we shall get the “second derived function” or second differential coefficient, which is denoted by ${\displaystyle f''(x)}$; and so on.

Now let us generalize.

Let ${\displaystyle y=f(x)=x^{n}}$.

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

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

if the original function be ${\displaystyle y=f(x)}$;

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

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

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

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

${\displaystyle {\frac {dy}{dx}}=f'(x)=28x^{3}+10.5x^{2}-x+1}$,

${\displaystyle {\frac {d^{2}y}{dx^{2}}}=f''(x)=84x^{2}+21x-1}$,

${\displaystyle {\frac {d^{3}y}{dx^{3}}}=f'''(x)=168x+21}$,

${\displaystyle {\frac {d^{4}y}{dx^{4}}}=f''''(x)=168}$,

${\displaystyle {\frac {d^{5}y}{dx^{5}}}=f'''''(x)=0}$.

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

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

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

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

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

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

(1) ${\displaystyle y=17x+12x^{2}}$.

(2) ${\displaystyle y={\frac {x^{2}+a}{x+a}}}$.

(3) ${\displaystyle y=1+{\frac {x}{1}}+{\frac {x^{2}}{1\times 2}}+{\frac {x^{3}}{1\times 2\times 3}}+{\frac {x^{4}}{1\times 2\times 3\times 4}}}$.

(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.