Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/50

Assuming $\scriptstyle {x=10}$ for the initial value of $\scriptstyle {x}$ fixes $\scriptstyle {y=100}$ as the initial value of $\scriptstyle {y}$.

Suppose	$\scriptstyle {x}$ increases to	$\scriptstyle {x}$	$\scriptstyle {=12}$,	that is,	$\scriptstyle {\Delta x}$	$\scriptstyle {=2}$;
then	$\scriptstyle {y}$ increases to	$\scriptstyle {y}$	$\scriptstyle {=144}$,	and	$\scriptstyle {\Delta y}$	$\scriptstyle {=44}$.
Suppose	$\scriptstyle {x}$ decreases to	$\scriptstyle {x}$	$\scriptstyle {=9}$,	that is,	$\scriptstyle {\Delta x}$	$\scriptstyle {=-1}$;
then	$\scriptstyle {y}$ increases to	$\scriptstyle {y}$	$\scriptstyle {=81}$,	and	$\scriptstyle {\Delta y}$	$\scriptstyle {=-19}$.

It may happen that as $\scriptstyle {x}$ increases, $\scriptstyle {y}$ decreases, or the reverse; in either case $\scriptstyle {\Delta x}$ and $\scriptstyle {\Delta y}$ will have opposite signs.

It is also clear (as illustrated in the above example) that if $\scriptstyle {y=f(x)}$ is a continuous function and $\scriptstyle {\Delta x}$ is decreasing in numerical value, then $\scriptstyle {\Delta y}$ also decreases in numerical value.

27. Comparison of increments. Consider the function

Assuming a fixed initial value for $\scriptstyle {x}$, let $\scriptstyle {x}$ take on an increment $\scriptstyle {\Delta x}$. Then $\scriptstyle {y}$ will take on a corresponding increment $\scriptstyle {\Delta y}$, and we have

	$\scriptstyle {y+}$	$\scriptstyle {\Delta y}$	$\scriptstyle {=(x+\Delta x)^{2}}$,	Subtracting (A)
or,	$\scriptstyle {y+}$	$\scriptstyle {\Delta y}$	$\scriptstyle {=x^{2}+2x\cdot \Delta x+(\Delta x)^{2}}$.
Subtracting (A),	$\scriptstyle {y}$		$\scriptstyle {=x^{2}}$
(B)		$\scriptstyle {\Delta y}$	$\scriptstyle {=}$$\scriptstyle {x^{2}+}$$\scriptstyle {2x\cdot \Delta x+(\Delta x)^{2}}$

we get the increment $\scriptstyle {\Delta y}$ in terms of $\scriptstyle {x}$ and $\scriptstyle {\Delta x}$.

To find the ratio of the increments, divide (B) by $\scriptstyle {\Delta x}$, giving$\scriptstyle {{\frac {\Delta y}{\Delta x}}=2x+\Delta x.}$

If the initial value of $\scriptstyle {x}$ is $\scriptstyle {4}$, it is evident that$\scriptstyle {{\underset {\Delta x=0}{\operatorname {limit} }}{\frac {\Delta y}{\Delta x}}=8.}$

Let us carefully note the behavior of the ratio of the increments of $\scriptstyle {x}$ and $\scriptstyle {y}$ as the increment of $\scriptstyle {x}$ diminishes.

Initial value of $\scriptstyle {x}$	New value of $\scriptstyle {x}$	Increment $\scriptstyle {\Delta x}$	Initial value of $\scriptstyle {y}$	New value of $\scriptstyle {y}$	Increment $\scriptstyle {\Delta y}$	$\scriptstyle {\frac {\Delta y}{\Delta x}}$
4	5.0	1.0	16	25.	9.	9.
4	4.8	0.8	16	23.04	7.04	8.8
4	4.6	0.6	16	21.16	5.16	8.6
4	4.4	0.4	16	19.36	3.36	8.4
4	4.2	0.2	16	17.64	1.64	8.2
4	4.1	0.1	16	16.81	0.81	8.1
4	4.01	0.01	16	16.0801	0.0801	8.01