Elements of the Differential and Integral Calculus/Chapter IV

by William Anthony Granville

CHAPTER IV

 

CHAPTER IV

 

DIFFERENTIATION

 

25. Introduction. We shall now proceed to investigate the manner in which a function changes in value as the independent variable changes. The fundamental problem of the Differential Calculus is to establish a measure of this change in the function with mathematical precision. It was while investigating problems of this sort, dealing with continuously varying quantities, that Newton^[1] was led to the discovery of the fundamental principles of the Calculus, the most scientific and powerful tool of the modern mathematician.

26. Increments. The increment of a variable in changing from one numerical value to another is the difference found by subtracting the first value from the second. An increment of $\scriptstyle {x}$ is denoted by the symbol $\scriptstyle {\Delta x}$, read delta $\scriptstyle {x}$.

The student is warned against reading this symbol delta times $\scriptstyle {x}$, it having no such meaning. Evidently this increment may be either positive or negative^[2] according as the variable in changing is increasing or decreasing in value. Similarly,

$\scriptstyle {\Delta y}$	denotes an increment of $\scriptstyle {y}$,
$\scriptstyle {\Delta \phi }$	denotes an increment of $\scriptstyle {\phi }$,
$\scriptstyle {\Delta f(x)}$	denotes an increment $\scriptstyle {f(x)}$, etc.

If in $\scriptstyle {y=f(x)}$ the independent variable $\scriptstyle {x}$, takes on an increment $\scriptstyle {\Delta x}$, then $\scriptstyle {\Delta y}$ is always understood to denote the corresponding increment of the function $\scriptstyle {f(x)}$ (or dependent variable $\scriptstyle {y}$).

The increment $\scriptstyle {\Delta y}$ is always assumed to be reckoned from a definite initial value of $\scriptstyle {y}$ corresponding to the arbitrarily fixed initial value of $\scriptstyle {x}$ from which the increment $\scriptstyle {\Delta x}$ is reckoned. For instance, consider the function$\scriptstyle {y=x^{2}}$.

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

It is apparent that as $\scriptstyle {\Delta x}$ decreases, $\scriptstyle {\Delta y}$ also diminishes, but their ratio takes on the successive values $\scriptstyle {9,~8.8,~8.6,~8.4,~8.2,~8.1,~8.01}$; illustrating the fact that $\scriptstyle {\frac {\Delta y}{\Delta x}}$ can be brought as near to $\scriptstyle {8}$ in value as we please by making $\scriptstyle {\Delta x}$ small enough. Therefore

$\scriptstyle {{\underset {\Delta x=0}{\operatorname {limit} }}\;{\frac {\Delta y}{\Delta x}}=8}$.^[3]

28. Derivative of a function of one variable. The fundamental definition of the Differential Calculus is:

The derivative^[4] of a function is the limit of the ratio of the increment of the function to the increment of the independent variable, when the latter increment varies and approaches the limit zero.

When the limit of this ratio exists, the function is said to be differentiable, or to possess a derivative.

The above definition may be given in a more compact form symbolically as follows: Given the function

and consider $\scriptstyle {x}$ to have a fixed value.

Let $\scriptstyle {x}$ take on an increment $\scriptstyle {\Delta x}$; then the function $\scriptstyle {y}$ takes on an increment $\scriptstyle {\Delta y}$, the new value of the function being

To find the increment of the function, subtract (A) from (B), giving

Dividing by the increment of the variable, $\scriptstyle {\Delta x}$, we get

The limit of this ratio when $\scriptstyle {\Delta x}$ approaches the limit zero is, from our definition, the derivative and is denoted by the symbol $\scriptstyle {\frac {dy}{dx}}$. Therefore

defines the derivative of $\scriptstyle {y}$ [or $\scriptstyle {f(x)}$] with respect to $\scriptstyle {x}$.

From (D) we also get:

$\scriptstyle {{\frac {dy}{dx}}={\underset {\Delta x=0}{\operatorname {limit} }}\;{\frac {\Delta y}{\Delta x}}}$.

The process of finding the derivative of a function is called differentiation.

It should be carefully noted that the derivative is the limit of the ratio, not the ratio of the limits. The latter ratio would assume the form $\scriptstyle {\frac {0}{0}}$, which is indeterminate (§14, p12).

29. Symbols for derivatives. Since $\scriptstyle {\Delta y}$ and $\scriptstyle {\Delta x}$ are always finite and have definite values, the expression$\scriptstyle {\frac {\Delta y}{\Delta x}}$is really a fraction. The symbol

$\scriptstyle {\frac {dy}{dx}}$,

however, is to be regarded not as a fraction but as the limiting value of a fraction. In many cases it will be seen that this symbol does possess fractional properties, and later on we shall show how meanings may be attached to $\scriptstyle {dy}$ and $\scriptstyle {dx}$, but for the present the symbol $\scriptstyle {\frac {dy}{dx}}$ is to be considered as a whole.

Since the derivative of a function of $\scriptstyle {x}$ is in general also a function of $\scriptstyle {x}$, the symbol $\scriptstyle {f'(x)}$ is also used to denote the derivative of $\scriptstyle {f(x)}$.

Hence, if	$\scriptstyle {y}$	$\scriptstyle {=f(x)}$,
we may write	$\scriptstyle {\frac {dy}{dx}}$	$\scriptstyle {=f'(x),}$

which is read the derivative of $\scriptstyle {y}$ with respect to $\scriptstyle {x}$ equals $\scriptstyle {f}$ prime of $\scriptstyle {x}$. The symbol$\scriptstyle {\frac {d}{dx}}$when considered by itself is called the differentiating operator, and indicates that any function written after it is to be differentiated with respect to $\scriptstyle {x}$. Thus

$\scriptstyle {\frac {dy}{dx}}$ or $\scriptstyle {{\frac {d}{dx}}y}$ indicates the derivative of $\scriptstyle {y}$ with respect to $\scriptstyle {x}$;
$\scriptstyle {{\frac {d}{dx}}f(x)}$ indicates the derivative of $\scriptstyle {f(x)}$ with respect to $\scriptstyle {x}$;
$\scriptstyle {{\frac {d}{dx}}(2x^{2}+5)}$ indicates the derivative of $\scriptstyle {2x^{2}+5}$ with respect to $\scriptstyle {x}$.
$\scriptstyle {y'}$ is an abbreviated form of $\scriptstyle {\frac {dy}{dx}}$.

The symbol $D_{x}$ is used by some writers instead of ${\frac {d}{dx}}$. If then

$y=f(x),$

we may write the identities

$y'={\frac {dy}{dx}}={\frac {d}{dx}}y={\frac {d}{dx}}f(x)=D_{x}f(x)=f'(x).$

30. Differentiable functions. From the Theory of Limits it is clear that if the derivative of a function exists for a certain value of the independent variable, the function itself must be continuous for that value of the variable.

The converse, however, is not always true, functions having been discovered that are continuous and yet possess no derivative. But such functions do not occur often in applied mathematics, and in this book only differentiable functions are considered, that is, functions that possess a derivative for all values of the independent variable save at most for isolated values.

31. General rule for differentiation. From the definition of a derivative it is seen that the process of differentiating a function $y=f(x)$ consists in taking the following distinct steps:

General Rule for Differentiation^[5]

First Step. In the function replace $x$ by $x+\Delta x$, giving a new value of the function, $y+\Delta y$.
Second Step. Subtract the given value of the function from the new value in order to find $\Delta y$ (the increment of the function).
Third Step. Divide the remainder $\Delta y$ (the increment of the function) by $\Delta x$ (the increment of the independent variable).
Fourth Step. Find the limit of this quotient, when $\Delta x$ (the increment of the independent variable) varies and approaches the limit zero. This is the derivative required.

The student should become thoroughly familiar with this rule by applying the process to a large number of examples. Three such examples will now be worked out in detail.

Illustrative Example 1. Differentiate $3x^{2}+5$.

Solution. Applying the successive steps in the General Rule, we get, after placing

	$y$	$=3x^{2}+5$,
First step.	$y+\Delta y$	$=3(x+\Delta x)^{2}+5$
		$=3x^{2}+6x\cdot \Delta x+3(\Delta x)^{2}+5$.
Second step.	$\scriptscriptstyle {y+\Delta y}$	$\scriptscriptstyle {=3x^{2}+6x\cdot \Delta x+3(\Delta x)^{2}}$	$\scriptscriptstyle {+5}$
	$\scriptscriptstyle {y}$	$\scriptscriptstyle {=3x^{2}}$	$\scriptscriptstyle {+5}$
	$\scriptscriptstyle {\Delta y}$	$\scriptscriptstyle {=6x\cdot \Delta x+3(\Delta x)^{2}}$.
Third step.	$\scriptscriptstyle {\frac {\Delta y}{\Delta x}}$	$\scriptscriptstyle {=6x+3\cdot \Delta x}$.
Fourth step.	$\scriptscriptstyle {\frac {dy}{dx}}$	$\scriptscriptstyle {=6x}$.   Ans.
We may also write this
$\scriptscriptstyle {{\frac {d}{dx}}(3x^{2}+5)}$		$\scriptscriptstyle {=6x}$.

 

Illustrative Example 2. Differentiate $\scriptscriptstyle {x^{3}-2x+7}$.

Solution.	Place	$\scriptscriptstyle {y}$		$\scriptscriptstyle {=x^{3}-2x+7}$.
First step.		$\scriptscriptstyle {y+\Delta y}$		$\scriptscriptstyle {=(x+\Delta x)^{3}-2(x+\Delta x)+7}$.
				$\scriptscriptstyle {=x^{3}}$	$\scriptscriptstyle {+3x^{2}\cdot \Delta x+3x\cdot (\Delta x)^{2}+(\Delta x)^{3}}$	$\scriptscriptstyle {-2x}$	$\scriptscriptstyle {-2\cdot \Delta x}$	$\scriptscriptstyle {+7}$.
Second step.			$\scriptscriptstyle {y+\Delta y}$	$\scriptscriptstyle {=x^{3}}$	$\scriptscriptstyle {+3x^{2}\cdot \Delta x+3x\cdot (\Delta x)^{2}+(\Delta x)^{3}}$	$\scriptscriptstyle {-2x}$	$\scriptscriptstyle {-2\cdot \Delta x}$	$\scriptscriptstyle {+7}$
			$\scriptscriptstyle {y}$$\scriptscriptstyle {+\Delta y}$	$\scriptscriptstyle {=x^{3}}$		$\scriptscriptstyle {-2x}$		$\scriptscriptstyle {+7}$
			$\scriptscriptstyle {\Delta y}$	$\scriptscriptstyle {=}$	$\scriptscriptstyle {3x^{2}\cdot \Delta x+3x\cdot (\Delta x)^{2}+(\Delta x)^{3}}$		$\scriptscriptstyle {-2\cdot \Delta x}$
Third step.		$\scriptscriptstyle {\frac {\Delta y}{\Delta x}}$		$\scriptscriptstyle {=3x^{2}+3x\cdot \Delta x+(\Delta x)^{2}-2.}$
Fourth step.		$\scriptscriptstyle {\frac {dy}{dx}}$		$\scriptscriptstyle {=3x^{2}-2}$.   Ans.
Or,	$\scriptscriptstyle {{\frac {d}{dx}}(x^{3}-2x+7)}$			$\scriptscriptstyle {=3x^{2}-2}$.

 

Illustrative Example 3. Differentiate $\scriptscriptstyle {\frac {c}{x^{2}}}$.

Solution.	Place	$\scriptscriptstyle {y}$	$\scriptscriptstyle {={\frac {c}{x^{2}}}}$.
First step.		$\scriptscriptstyle {y+\Delta y}$	$\scriptscriptstyle {={\frac {c}{(x+\Delta x)^{2}}}}$.
Second step.		$\scriptscriptstyle {y+\Delta y}$	$\scriptscriptstyle {={\frac {c}{(x+\Delta x)^{2}}}}$
		$\scriptscriptstyle {y}$$\scriptscriptstyle {+\Delta y}$	$\scriptscriptstyle {={\frac {c}{x^{2}}}}$
$\scriptscriptstyle {\Delta y}$			$\scriptscriptstyle {={\frac {c}{(x+\Delta x)^{2}}}-{\frac {c}{x^{2}}}={\frac {-c\cdot \Delta x(2x+\Delta x)}{x^{2}(x+\Delta x)^{2}}}}$.
Third step.		$\scriptscriptstyle {\frac {\Delta y}{\Delta x}}$	$\scriptscriptstyle {=-c\cdot {\frac {2x+\Delta x}{x^{2}(x+\Delta x)^{2}}}}$.
Fourth step.		$\scriptscriptstyle {\frac {dy}{dx}}$	$\scriptscriptstyle {=-c\cdot {\frac {2x}{x^{2}(x)^{2}}}}$.
			$\scriptscriptstyle {=-{\frac {2c}{x^{3}}}}$.   Ans.
Or,	$\scriptscriptstyle {{\frac {d}{dx}}\left({\frac {c}{x^{2}}}\right)}$		$\scriptscriptstyle {={\frac {-2c}{x^{3}}}}$.

EXAMPLES

Use the General Rule, p. 291, in differentiating the following functions:

 1. $\scriptstyle {y=3x^{2}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=6x.}$
 2. $\scriptstyle {y=x^{2}+2.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=2x.}$
 3. $\scriptstyle {y=5-4x.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=-4.}$
 4. $\scriptstyle {s=2t^{2}-4.}$Ans. $\scriptstyle {{\frac {ds}{dt}}=4t.}$
 5. $\scriptstyle {y={\frac {1}{x}}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=-{\frac {1}{x^{2}}}.}$
 6. $\scriptstyle {y={\frac {x+2}{x}}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=-{\frac {2}{x^{2}}}.}$
 7. $\scriptstyle {y=x^{3}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=3x^{2}.}$
 8. $\scriptstyle {y=2x^{2}-3.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=4x.}$
 9. $\scriptstyle {y=1-2x^{3}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=4x.}$
10. $\scriptstyle {\rho =a\theta ^{2}.}$Ans. $\scriptstyle {{\frac {d\rho }{d\theta }}=2a\theta .}$
11. $\scriptstyle {y={\frac {2}{x^{2}}}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=-{\frac {4}{x^{3}}}.}$
12. $\scriptstyle {y={\frac {3}{x-1}}.}$Ans. $\scriptstyle {{\frac {dy}{dx}}=-{\frac {3}{(x-1)^{2}}}.}$
13. $\scriptstyle {y=7x^{2}+x.}$
14. $\scriptstyle {s=at^{2}-2bt.}$
15. $\scriptstyle {r=8t+3t^{2}.}$
16. $\scriptstyle {y={\frac {3}{x^{2}}}.}$
17. $\scriptstyle {s=-{\frac {a}{2t+3}}.}$
18. $\scriptstyle {y=bx^{3}-cx.}$
19. $\scriptstyle {\rho =3\theta ^{3}-2\theta ^{2}.}$
20. $\scriptstyle {y={\frac {3}{4}}x^{2}-{\frac {1}{2}}x.}$
21. $\scriptstyle {y={\frac {x^{2}-5}{x}}.}$
22. $\scriptstyle {\rho ={\frac {\theta ^{2}}{1+\theta }}.}$
23. $\scriptstyle {y={\frac {1}{2}}x^{2}+2x.}$
24. $\scriptstyle {z=4x-3x^{2}.}$
25. $\scriptstyle {\rho =3\theta +\theta ^{2}.}$
26. $\scriptstyle {y={\frac {ax+b}{x^{2}}}.}$
27. $\scriptstyle {z={\frac {x^{3}+2}{x}}.}$
28. $\scriptstyle {y=x^{2}-3x+6.}$Ans. $\scriptstyle {y'=2x-3.}$
29. $\scriptstyle {s=2t^{2}+5t-8.}$Ans. $\scriptstyle {s'=4t+5.}$.
30. $\scriptstyle {\rho =5\theta ^{3}-2\theta +6.}$Ans. $\scriptstyle {\rho '=15\theta ^{2}-2.}$
31. $\scriptstyle {y=ax^{2}+bx+c.}$Ans. $\scriptstyle {y'=2ax+b.}$

32. Applications of the derivative to Geometry. We shall now consider a theorem which is fundamental in all applications of the Differential Calculus to Geometry. Let

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

be the equation of a curve $\scriptstyle {AB}$.

Now differentiate (A) by the General Rule and interpret each step geometrically.

First Step.	$\scriptstyle {y+\Delta y}$	$\scriptstyle {=f(x+\Delta x)}$	$\scriptstyle {=NQ}$
Second Step.	$\scriptstyle {y+\Delta y}$	$\scriptstyle {=f(x+\Delta x)}$	$\scriptstyle {=NQ}$
	$\scriptstyle {y}$$\scriptstyle {+\Delta y}$	$\scriptstyle {=f(x)}$	$\scriptstyle {=MP=NR}$
	$\scriptstyle {\Delta y}$	$\scriptstyle {=f(x+\Delta x)-f(x)}$	$\scriptstyle {=RQ.}$

Third Step.	$\scriptstyle {\frac {\Delta y}{\Delta x}}$	$\scriptstyle {={\frac {f(x+\Delta x)-f(x)}{\Delta x}}={\frac {RQ}{MN}}={\frac {RQ}{PR}}}$
		$\scriptstyle {=\tan RPQ=\tan \phi }$
		=slope of secant line $\scriptstyle {PQ.}$

Fourth Step.	$\lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$	$=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$
(B)		$={\frac {dy}{dx}}=$ value of the derivative at $P$.

But when we let $\Delta x{\dot {=}}0$, the point $Q$ will move along the curve and approach nearer and nearer to $P$, the secant will turn about $P$ and approach the tangent as a limiting position, and we have also

	$\lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$	$=\lim _{\Delta x\to 0}\tan \phi =\tan \tau$
(C)		= slope of the tangent at $P$.

Hence from (B) and (C), ${\tfrac {dy}{dx}}=$ slope of the tangent line $PT$. Therefore

Theorem. The value of the derivative at any point of a curve is equal to the slope of the line drawn tangent to the curve at that point.

It was this tangent problem that led Leibnitz^[6] to the discovery of the Differential Calculus.

Illustrative Example 1. Find the slopes of the tangents to the parabola $y=x^{2}$ at the vertex, and at the point where $x={\tfrac {1}{2}}$.

Solution. Differentiating by General Rule, §31, we get

(A)	${\frac {dy}{dx}}=2x=$ slope of tangent line at any point on curve.

Tangent to a parabola.

To find slope of tangent at vertex, substitute $x=0$ in (A), giving

${\frac {dy}{dx}}=0$.

Therefore the tangent at vertex has the slope zero; that is, it is parallel to the axis of x and in this case coincides with it.

To find slope of tangent at the point $P$, where $x={\tfrac {1}{2}}$, substitute in (A), giving

${\frac {dy}{dx}}=1$;

that is, the tangent at the point $P$ makes an angle of 45° with the axis of $x$.

EXAMPLES

Find by differentiation the slopes of the tangents to the following curves at the points indicated. Verify each result by drawing the curve and its tangent.

1. $y=x^{2}-4$,	where $x=2$.	Ans. 4.
2. $y=6-3x^{2}$	where $x=1$.	-6.
3. $y=x^{3}$,	where $x=-1$.	-3.
4. $y={\frac {2}{x}}$,	where $x=-1$.	$-{\frac {1}{2}}$.
5. $y=x-x^{2}$,	where $x=0$.	1.
6. $y={\frac {1}{x-1}}$,	where $x=3$.	$-{\frac {1}{4}}$.
7. $y={\frac {1}{2}}x^{2}$,	where $x=4$.	4.
8. $y=x^{2}-2x+3$,	where $x=1$.	0.
9. $y=9-x^{2}$,	where $x=-3$.	6.

10. Find the slope of the tangent to the curve $y=2x^{3}-6x+5$, (a) at the point where $x=1$; (b) at the point where $x=0$.

Ans. (a) 0; (b) -6.

11. (a) Find the slopes of the tangents to the two curves $y=3x^{2}-1$ and $y=2x^{2}+3$ at their points of intersection. (b) At what angle do they intersect?

Ans. (a) $\pm 12,\pm 8$; (b) $\arctan {\frac {4}{97}}$.

12. The curves on a railway track are often made parabolic in form. Suppose that a track has the form of the parabola $y=x^{2}$ (last figure, § 32), the directions $OX$ and $OY$ being east and north respectively, and the unit of measurement 1 mile. If the train is going east when passing through $O$, in what direction will it be going

(a) when ${\tfrac {1}{2}}$ mi. east of $OY$?	Ans. Northeast.
(b) when ${\tfrac {1}{2}}$ mi. west of $OY$?	Southeast.
(c) when ${\tfrac {\sqrt {3}}{2}}$ mi. east of $OY$?	N. 30°E.
(d) when ${\tfrac {1}{12}}$ mi. north of $OX$?	E. 30°S., or E. 30°N.

13. A street-car track has the form of the cubical parabola $y=x^{3}$. Assume the same directions and unit as in the last example. If a car is going west when passing through $O$, in what direction will it be going

(a) when ${\tfrac {1}{\sqrt {3}}}$ mi. east of $OY$?	Ans. Southwest.
(b) when ${\tfrac {1}{\sqrt {3}}}$ mi. west of $OY$?	Southwest.
(c) when ${\tfrac {1}{2}}$ mi. north of $OX$?	S. 27° 43' W.
(d) when 2 mi. south of $OX$?
(e) when equidistant from $OX$ and $OY$?

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1. Sir Isaac Newton (1642–1727), an Englishman, was a man of the most extraordinary genius. He developed the science of the Calculus under the name of Fluxions. Although Newton had discovered and made use of the new science as early as 1670, his first published work in which it occurs is dated 1687, having the title Philosophiae Naturalis Principia Mathematica. This was Newton's principal work. Laplace said of it, "It will always remain preëminent above all other productions of the human mind." See frontispiece.
2. Some writers call a negative increment a decrement.
3. The student should guard against the common error of concluding that because the numerator and denominator of a fraction are each approaching zero as a limit, the limit of the value of the fraction (or ratio) is zero. The limit of the ratio may take on any numerical value. In the above example the limit is $\scriptstyle {8}$.
4. Also called the differential coefficient or the derived function.
5. Also called the Four-step Rule.
6. Gottfried Wilhelm Leibnitz (1646-1716) was a native of Leipzig. His remarkable abilities were shown by original investigations in several branches of learning. He was first to publish his discoveries in Calculus in a short essay appearing in the periodical Acta Eruditorum at Leipzig in 1684. It is known, however, that manuscripts on Fluxions written by Newton were already in existence, and from these some claim Leibnitz got the new ideas. The decision of modern times seems to be that both Newton and Leibnitz invented the Calculus independently of each other. The notation nsed to-day was introduced by Leibnitz. See frontispiece.

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