Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/163

93. Successive differentials. As the differential of a function is in general also a function of the independent variable, we may deal with its differential. Consider the function

$\scriptstyle {y=f(x)}$.

$\scriptstyle {d(dy)}$ is called the second differential of $\scriptstyle {y}$ (or of the function) and is denoted by the symbol$\scriptstyle {d^{2}y.}$

Similarly, the third differential of $\scriptstyle {y}$, $\scriptstyle {d[d(dy)]}$, is written

$\scriptstyle {d^{3}y}$,

and so on, to the nth differential of $\scriptstyle {y}$,

$\scriptstyle {d^{n}y}$.

Since $\scriptstyle {dx}$, the differential of the independent variable, is independent of $\scriptstyle {x}$ (see footnote , p. 131), it must be treated as a constant when differentiating with respect to $\scriptstyle {x}$. Bearing this in mind, we get very simple relations between successive differentials and successive derivatives. For ${\begin{aligned}\scriptstyle {dy}&\scriptstyle {=f'(x)dx,}\\\scriptstyle {d^{2}y}&\scriptstyle {=f''(x)(dx)^{2},}\end{aligned}}$
and
since dx is regarded as a constant.

Also,${\begin{aligned}\scriptstyle {d^{3}y}&\scriptstyle {=f'''(x)(dx)^{3},}\\\scriptstyle {d^{n}y}&\scriptstyle {=f^{(n)}(x)(dx)^{n}.}\end{aligned}}$
and in general

Dividing both sides of each expression by the power of $\scriptstyle {dx}$ occurring on the right, we get our ordinary derivative notation

$\scriptstyle {{\frac {d^{2}y}{dx^{2}}}=f''(x),\quad {\frac {d^{3}y}{dx^{3}}}=f'''(x),\quad \dots ,\quad {\frac {d^{n}y}{dx^{n}}}=f^{(n)}(x)}$.

Powers of an infinitesimal are called infinitesimals of a higher order. More generally, if for the infinitesimals $\scriptstyle {\alpha }$ and $\scriptstyle {\beta }$,

$\scriptstyle {\operatorname {limit\;} {\frac {\beta }{\alpha }}=0}$,

then $\scriptstyle {\beta }$ is said to be an infinitesimal of a higher order than $\scriptstyle {\alpha }$.

Illustrative Example 1. Find the third differential of

$\scriptscriptstyle {y=x^{5}-2x^{3}+3x-5}$.

Solution.${\begin{aligned}\scriptscriptstyle {dy}&\scriptscriptstyle {=(5x^{4}-6x^{2}+3)dx,}\\\scriptscriptstyle {d^{2}y}&\scriptscriptstyle {=(20x^{3}-12x)(dx)^{2},}\\\scriptstyle {d^{3}y}&\scriptscriptstyle {=(60x^{2}-12)(dx)^{3}.\qquad {\mathit {Ans.}}}\end{aligned}}$

Note. This is evidently the third derivative of the function multiplied by the cube of the differential of the independent variable. Dividing through by $\scriptscriptstyle {(dx)^{3}}$, we get the third derivative

$\scriptscriptstyle {{\frac {d^{3}y}{dx^{3}}}=60x^{2}-12}$.