Elements of the Differential and Integral Calculus/Chapter IX

by William Anthony Granville

CHAPTER IX

DIFFERENTIALS

87. Introduction. Thus far we have represented the derivative of $y=f(x)$ by the notation

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

We have taken special pains to impress on the student that the symbol

${\frac {dy}{dx}}$

was to be considered not as an ordinary fraction with dy as numerator and dx as denominator, but as a single symbol denoting the limit of the quotient

${\frac {\Delta y}{\Delta x}}$

as $\Delta x$ approaches the limit zero.

Problems do occur, however, where it is very convenient to be able to give a meaning to dx and dy separately, and it is especially useful in applications of the Integral Calculus. How this may be done is explained in what follows.

88. Definitions. If $f'(x)$ is the derivative of $f(x)$ for a particular value of x, and $\Delta x$ is an arbitrarily chosen increment of x, then the differential of $f(x)$, denoted by the symbol $df(x)$, is defined by the equation

(A)	$df(x)=f'(x)\Delta x.$

If now $f(x)=x$, then $f'(x)=1$, and (A) reduces to

$dx=\Delta x,$

showing that when x is the independent variable, the differential of x (= dx) is identical with $\Delta x$. Hence, if $y=f(x)$, (A) may in general be written in the form

(B)	$dy=f'(x)dx.$[1]

The differential of a function equals its derivative multiplied by the differential of the independent variable.

Let us illustrate what this means geometrically.

Graph of function.

Let $f'(x)$ be the derivative of $y=f(x)$ at P. Take $dx=PQ$, then

$dy=f'(x)dx=\tan \tau \cdot PQ={\frac {QT}{PQ}}\cdot PQ=QT.$

Therefore $dy$, or $df(x)$, is the increment (= QT) of the ordinate of the tangent corresponding to dx.^[2]

This gives the following interpretation of the derivative as a fraction.

If an arbitrarily chosen increment of the independent variable x for a point P(x, y) on the curve y = f(x) be denoted by dx, then in the derivative

${\frac {dy}{dx}}=f'(x)=\tan \tau ,$

dy denotes the corresponding increment of the ordinate drawn to the tangent.

89. Infinitesimals. In the Differential Calculus we are usually concerned with the derivative, that is, with the ratio of the differentials dy and dx. In some applications it is also useful to consider dx as an infinitesimal (see §15), that is, as a variable whose values remain numerically small, and which, at some stage of the investigation, approaches the limit zero. Then by (B), §88, and (2), §20, dy is also an infinitesimal.

In problems where several infinitesimals enter we often make use of the following

Theorem. In problems involving the limit of the ratio of two infinitesimals, either infinitesimal may be replaced by an infinitesimal so related to it that the limit of their ratio is unity.

Proof. Let $\alpha ,\beta ,\alpha ',\beta '$ be infinitesimals so related that

(C)	limit ${\frac {\alpha '}{\alpha }}=1$ and limit ${\frac {\beta '}{\beta }}=1$.

We have	${\frac {\alpha }{\beta }}$	$={\frac {\alpha '}{\beta '}}\cdot {\frac {alpha}{\alpha '}}\cdot {\frac {\beta '}{\beta }}$ identically,
and	$\lim {\frac {\alpha }{\beta }}$	$=\lim {\frac {\alpha '}{\beta '}}\cdot \lim {\frac {\alpha }{\alpha '}}\cdot \lim {\frac {\beta '}{\beta }}$	Th. II, §20
		$=\lim {\alpha '}{\beta '}\cdot 1\cdot 1$.	By (C)
(D)	∴ $\lim {\frac {\alpha }{\beta }}$	$=\lim {\frac {\alpha '}{\beta '}}$	Q.E.D

Now let us apply this theorem to the two following important limits.

For the independent variable x, we know from the previous section that $\Delta x$ and dx are identical.

Hence their ratio is unity, and also limit ${\tfrac {\Delta x}{dx}}=1$. That is, by the above theorem,

(E)	In the limit of the ratio of $\Delta x$ and a second infinitesimal, $\Delta x$ may be replaced by dx.

On the contrary it was shown that, for the dependent variable y, $\Delta y$ and dy are in general unequal. But we shall now show, however, that in this case also

$\lim {\frac {\Delta y}{dy}}=1$.

Since $\lim _{\Delta x=0}{\tfrac {\Delta y}{\Delta x}}=f'(x)$ we may write

${\frac {\Delta y}{\Delta x}}=f'(x)+\epsilon$,

where $\epsilon$ is an infinitesimal which approaches zero when $\Delta x{\dot {=}}0$.

Clearing of fractions, remembering that $\Delta x{\dot {=}}dx$,

	$\Delta y$	$=f'(x)dx+\epsilon \cdot \Delta x$,
or	$\Delta y$	$=dy+\epsilon \cdot \Delta x$,	(B), §89

Dividing both sides by $\Delta y$,

	$1$	$={\frac {dy}{\Delta y}}+\epsilon \cdot {\frac {\Delta x}{\Delta y}}$,
or	${\frac {dy}{\Delta y}}$	$=1-\epsilon \cdot {\frac {\Delta x}{\Delta y}}$.
		∴ $\lim _{\Delta x\to 0}{\frac {dy}{\Delta y}}=1$,

and hence $\lim _{\Delta x\to 0}{\tfrac {\Delta y}{dy}}=1$. That is, by the above theorem,

(F)	In the limit of the ratio of $\Delta y$ and a second infinitesimal, $\Delta y$ may be replaced by dy.

90. Derivative of the arc in rectangular coördinates. Let s be the length^[3] of the arc AP measured from a fixed point A on the curve.

Denote the increment of s (= arc PQ) by $\Delta s$. The definition of the length of arc depends on the assumption that, as Q approaches P,

$\lim \left({\frac {{\mbox{chord}}PQ}{{\mbox{arc}}PQ}}\right)=1.$

If we now apply the theorem in §89 to this, we get

(G)	In the limit of the ratio of chord PQ and a second infinitesimal, chord PQ may be replaced by arc PQ (= $\Delta s$).

From the above figure

(H)	$({\mbox{chord}}PQ)^{2}=(\Delta x)^{2}+(\Delta y)^{2},$

Dividing through by $(\Delta x)^{2}$, we get

(I)	$\left({\frac {{\mbox{chord}}PQ}{\Delta x}}\right)^{2}=1+\left({\frac {\Delta y}{\Delta x}}\right)^{2}$.

Now let Q approach P as a limiting position; then $\Delta x{\dot {=}}0$ and we have

$\left({\frac {ds}{dx}}\right)^{2}=1+\left({\frac {dy}{dx}}\right)^{2}$.

[Since $\lim _{\Delta x\to 0}\left({\tfrac {chordPQ}{\Delta x}}\right)=\lim _{\Delta x\to 0}\left({\tfrac {\Delta s}{\Delta x}}\right)={\tfrac {ds}{dx}}$, (G).]

(24)	∴ ${\frac {ds}{dx}}={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}.$

Similarly, if we divide (H) by $(\Delta y)^{2}$ and pass to the limit, we get

(25)	${\frac {ds}{dy}}={\sqrt {\left({\frac {dx}{dy}}\right)^{2}+1}}.$

Also, from the above figure,

$\cos \theta ={\frac {\Delta x}{{\mbox{chord}}PQ}}$ $\sin \theta ={\frac {\Delta y}{{\mbox{chord}}PQ}}.$

Now as Q approaches P as a limiting position $\theta {\dot {=}}\tau$, and we get

(26)	$\cos \tau ={\frac {dx}{ds}}$, $\sin \tau ={\frac {dy}{ds}}.$

[Since from (G) $\lim {\tfrac {\Delta x}{{\mbox{chord}}PQ}}=\lim {\tfrac {\Delta x}{\Delta x}}={\tfrac {dx}{ds}}$, and $\lim {\tfrac {\Delta y}{{\mbox{chord}}PQ}}=\lim {\tfrac {\Delta y}{\Delta s}}={\tfrac {dy}{ds}}$.]

Using the notation of differentials, formulas (25) and (26) may be written

(27)	$ds=\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}dx$.
(28)	$ds=\left[\left({\frac {dx}{dy}}\right)^{2}+1\right]^{\frac {1}{2}}dy$.

Substituting the value of ds from (27) in (26),

(29)	$\cos \tau ={\frac {1}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}}},\ \sin \tau ={\frac {\frac {dy}{dx}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}}}$

An easy way to remember the relations (24)-(26) differentials dx, dy, ds is to note that they are correctly represented by a right triangle whose hypotenuse is ds, whose sides are dx and dy, and whose angle at the base is $\tau$. Then

$ds={\sqrt {(dx)^{2}+(dy)^{2}}},$

and, dividing by dx or dy, gives (24) or (25) respectively. Also, from the figure,

$\cos \tau ={\frac {dx}{ds}},\ \sin \tau ={\frac {dy}{ds}};$

the same relations given by (26).

91. Derivative of the arc in polar coördinates. In the derivation which follows we shall employ the same figure and the same notation used in §67

From the right triangle PRQ

$({\mbox{chord}}PQY)^{2}$	$=(PR)^{2}+(RQ)^{2}$
	$=(\rho \sin \Delta \theta )^{2}+(\rho +\Delta \rho -\rho \cos \Delta \theta )^{2}$.

Dividing throughout by $(\Delta \theta )^{2}$, we get

$\left({\frac {{\mbox{chord}}PQ}{\Delta \theta }}\right)^{2}=\rho ^{2}\left({\frac {\sin \Delta \theta }{\Delta \theta }}\right)^{2}+\left({\frac {\Delta \rho }{\Delta \theta }}+\rho \cdot {\frac {1-\cos \Delta \theta }{\Delta \theta }}\right)^{2}.$

Passing to the limit as $\Delta \theta$ diminishes towards zero, we get^[4]

	$\left({\frac {ds}{d\theta }}\right)^{2}=\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}$.
(30)	∴ ${\frac {ds}{d\theta }}={\sqrt {\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}}}$.
In the notation of differentials this becomes
(31)	$\ ds=\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {1}{2}}d\theta$.

These relations between $\rho$ and the differentials ds, dp, and $d\theta$ are correctly represented by a right triangle whose hypotenuse is ds and whose sides are $d\rho$ and $\rho d\theta$. Then

$ds={\sqrt {(\rho d\theta )^{2}+(d\rho )^{2}}},$

and dividing by $d\theta$ gives (30).

Denoting by $\psi$ the angle between $dp$ and $ds$, we get at once

$\tan \psi =\rho {\frac {d\theta }{d\rho }},$

which is the same as (A), §67.

Illustrative Example 1. Find the differential of the arc of the circle $x^{2}+y^{2}=r^{2}$.

Solution. Differentiating, ${\tfrac {dy}{dx}}=-{\tfrac {x}{y}}$.

To find ds in terms of x we substitute in (27), giving

$ds=\left[1+{\frac {x^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx=\left[{\frac {y^{2}+x^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx=\left[{\frac {r^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx={\frac {rdy}{\sqrt {r^{2}-x^{2}}}}.$

To find ds in terms of y we substitute in (28), giving

$ds=\left[1+{\frac {y^{2}}{x^{2}}}\right]^{\frac {1}{2}}dy=\left[{\frac {x^{2}+y^{2}}{x^{2}}}\right]^{\frac {1}{2}}dy=\left[{\frac {r^{2}}{x^{2}}}\right]^{\frac {1}{2}}={\frac {rdy}{\sqrt {r^{2}-y^{2}}}}.$

Illustrative Example 2. Find the differential of the arc of the cardioid $\rho =a(l-\cos \theta )$ in terms of $\theta$.

Solution. Differentiating, ${\tfrac {d\rho }{d\theta }}=a\sin \theta$

Substituting in (31), gives

$ds=[a^{2}(1-\cos \theta )^{2}+a^{2}\sin ^{2}\theta ]^{\frac {1}{2}}d\theta =a[2-2\cos \theta ]^{\frac {1}{2}}d\theta =a\left[4\sin ^{2}{\frac {\theta }{2}}\right]^{\frac {1}{2}}d\theta =2a\sin {\frac {\theta }{2}}d\theta .$

EXAMPLES

Find the differential of arc in each of the following curves:

1. $y^{2}=4x$.	Ans.	$ds={\sqrt {\frac {1+x}{x}}}dx.$
2. $y=ax^{2}$.		$ds={\sqrt {1+4a^{2}x^{2}}}dx$.
3. $y=x^{3}$.		$ds={\sqrt {1+9x^{4}}}dx$.
4. $y^{3}=x^{2}$.		$ds={\frac {1}{2}}{\sqrt {4+9y}}dy$.
5. $x^{\frac {2}{3}}+y^{\frac {2}{3}}=a^{\frac {2}{3}}$.		$ds={\sqrt[{3}]{\frac {a}{y}}}dy$.
6. $b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}$.		$ds={\sqrt {\frac {a^{2}-e^{2}x^{2}}{a^{2}-x^{2}}}}dx$.
HINT. $e^{2}={\frac {a^{2}-b^{2}}{a^{2}}}$.
7. $e^{y}\cos x=1$.		$ds=\sec \ x\ dx$
8. $\rho =a\cos \theta$.		$ds=a\ d\theta$
9. $\rho ^{2}=a^{2}\cos 2\theta$.		$ds={\sqrt {\sec 2\theta }}d\theta$.
10. $\rho =ae^{\theta \cot \alpha }$.		$ds=\rho \csc \alpha d\theta$.
11. $\rho =a^{\theta }$.		$ds=a^{\theta }{\sqrt {1+\log ^{2}a}}d\theta$.
12. $\rho =a\theta$.		$ds={\frac {1}{a}}{\sqrt {a^{2}+\rho ^{2}}}d\rho$.

13.	(a) $x^{2}-y^{2}=a^{2}$.	(h) $x^{\frac {1}{2}}+y^{\frac {1}{2}}=a^{\frac {1}{2}}$.
	(b) $x^{2}=4ay$.	(i) $y^{2}=ax^{3}$.
	(c) $y=e^{x}+e^{-x}$.	(j) $y=\log x$.
	(d) $xy=a$.	(k) $4x=y^{3}$.
	(e) $y=\log \sec x$.	(l) $\rho =a\sec ^{2}{\frac {\theta }{2}}$.
	(f) $\rho =2a\tan \theta \sin \theta$.	(m) $\rho =1+\sin \theta$.
	(g) $\rho =a\sec ^{3}{\frac {\theta }{3}}$.	(n) $\rho \theta =a$.

92. Formulas for finding the differentials of functions. Since the differential of a function is its derivative multiplied by the differential of the independent variable, it follows at once that the formulas for finding differentials are the same as those for finding derivatives given in § 33, if we multiply each one by dx.

This gives us

I		$d(c)$	$=0$.
II		$d(x)$	$=dx$.
III		$d(u+v-w)$	$=du+dv-dw$.
IV		$d(cv)$	$=cdv$.
V		$d(uv)$	$=udv+vdu$.
VI		$d(v^{n})$	$=nv^{n-1}dv.$

VI	a	$d(x^{n})$	$=nx^{n-1}dx$.
VII		$d\left({\frac {u}{v}}\right)$	$={\frac {vdu-udv}{v^{2}}}$.
VII	a	$d\left({\frac {u}{c}}\right)$	$={\frac {du}{c}}$.
VIII		$d(log_{a}v)$	$=\log _{a}e{\frac {dv}{v}}$.
IX		$d(a^{v})$	$=a^{v}\log adv$.
IX	a	$d(e^{v})=$	$e^{v}dv$.
X		$d(u^{v})$	$=vu^{v-1}du+\log u\cdot u^{v}\cdot dv$.
XI		$d(\sin v)$	$=\cos vdv$.
XII		$d(\cos v)$	$=-\sin vdv$.
XIII		$d(\tan v)$	$=\sec ^{2}vdv$, etc.
XVIII		$d(\arcsin v)$	$={\frac {dv}{\sqrt {1-v^{2}}}}$, etc.

The term "differentiation" also includes the operation of finding differentials.

In finding differentials the easiest way is to find the derivative as usual, and then multiply the result by dx.

Illustrative Example 1. Find the differential of

	$\ y$	$={\frac {x+3}{x^{2}+3}}$
Solution.	$dy=d\left({\frac {x+3}{x^{2}+3}}\right)$	$={\frac {(x^{2}+3)d(x+3)-(x+3)d(x^{2}+3)}{(x^{2}+3)^{2}}}$
	$={\frac {(x^{2}+3)dx-(x+3)2xdx}{(x^{2}+3)^{2}}}$	$={\frac {(3-6x-x^{2})dx}{(x^{2}+3)^{2}}}$, Ans.

Illustrative Example 2. Find dy from

	$b^{2}x^{2}-a^{2}y^{2}$	$a^{2}b^{2}$
Solution.	$2b^{2}xdx-2a^{2}ydy$	$0$.
	∴ $dy$	$={\frac {b^{2}x}{a^{2}y}}dx$. Ans.

Illustrative Example 3. Find dy from

	$\rho ^{2}$	$=a^{2}\cos s\theta$
Solution.	$2\rho d\rho$	$=-a^{2}\sin 2\theta \cdot 2d\theta$.
	∴ $d\rho$	$=-{\frac {a^{2}\sin 2\theta }{\rho }}d\theta$.

Illustrative Example 4. Find $d[\arcsin(3t-4t^{3})]$.

Solution. $d[\arcsin(3t-4t^{3})]={\frac {d(3t-4t^{3})}{\sqrt {1-(3t-4t^{3})^{2}}}}={\frac {3dt}{\sqrt {1-t^{2}}}}$. Ans. 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

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

EXAMPLES

Differentiate the following, using differentials:

1. $y=ax^{3}-bx^{2}+cx+d$.	Ans.	$dy=(3ax^{2}-2bx+c)dx$.
2. $y=2x^{\frac {5}{2}}-3x^{\frac {2}{3}}+6x^{-1}+5$.		$dy=(5x^{\frac {3}{2}}-2x^{-{\frac {1}{3}}-6x^{-2}})dx$
3. $y=(a^{2}-x^{2})^{5}$.		$dy=-10x(a^{2}-x^{2})^{4}dx$.
4. $y={\sqrt {1+x^{2}}}$		$dy={\frac {x}{\sqrt {1+x^{2}}}}dx$.
5. $y={\frac {x^{2n}}{(1+x^{2})^{n}}}$.		$dy={\frac {2nx^{2n-1}}{(1+x^{2})^{n+1}}}dx$.
6. $y=\log {\sqrt {1-x^{3}}}$.		$dy={\frac {3x^{2}dx}{2(x^{3}-1)}}$.
7. $y=(e^{x}+e^{-x})^{2}$.		$dy=2(e^{2x}-e^{-2x})dx$.
8. $y=e^{x}\log x$.		$dy=e^{x}\left(\log x+{\frac {1}{x}}\right)dx$.
9. $s=t-{\frac {e^{t}-e^{-t}}{e^{t}+e^{-t}}}$.		$ds=\left({\frac {e^{t}-e^{-t}}{e^{t}+e^{-t}}}\right)^{2}dt$.
10. $\rho =\tan \psi +\sec \psi$		$d\rho ={\frac {1+\sin \psi }{\cos ^{2}\psi }}d\psi$.
11. $r={\frac {1}{3}}\tan ^{3}\theta \tan \theta$.		$dr=\sec ^{4}\theta d\theta$.
12. $f(x)=(\log x)^{3}$.		$f'(x)dx={\frac {3(\log x)^{2}dx}{x}}$.
13. $\psi (t)={\frac {t^{3}}{(1-t^{2}}}^{\frac {3}{2}}$.		$\psi '(t)dt={\frac {3t^{2}dt}{(1-t^{2})^{\frac {5}{2}}}}$.

14. $d\left[{\frac {x\log x}{1-x}}+\log(1-x)\right]={\frac {\log xdx}{(1-x)^{2}}}$.

15. $d[\arctan \log y]={\frac {dy}{y[1+(\log y)^{2}]}}$.

16. $d\left[r\operatorname {arcvers} {\frac {y}{r}}-{\sqrt {2ry-y^{2}}}\right]={\frac {ydy}{\sqrt {2ry-y^{2}}}}$.

17. $d\left[{\frac {\cos \psi }{2\sin ^{2}\psi }}-{\frac {1}{2}}\log \tan {\frac {\psi }{2}}\right]=-{\frac {d\psi }{\sin ^{3}\psi }}$.

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1. On account of the position which the derivative $f'(x)$ here occupies, it is sometimes called the differential coefficient.

   The student should observe the important fact that, since dx may be given any arbitrary value whatever, dx is independent of x. Hence, dy is a function of two independent variables x and dx.

2. The student should note especially that the differential (= dy) and the increment (= dy) of the function corresponding to the same value of dx ($=\Delta x$) are not in general equal. For, in the figure, $dy=QT$, but $\Delta y=QP'$.
3. Defined in § 209.

4. $\lim _{\Delta \theta \to 0}{\frac {{\mbox{chord}}PQ}{\Delta \theta }}=\lim _{\Delta \theta \to 0}{\frac {\Delta s}{\Delta \theta }}={\frac {ds}{d\theta }}$	By (G), §90
   $\lim _{\Delta \to 0}{\frac {\sin \Delta \theta }{\Delta \theta }}=1$.	By §22
   $\lim _{\Delta \theta \to 0}{\frac {1-\cos \Delta \theta }{\Delta \theta }}=\lim _{\Delta \theta \to 0}{\frac {2\sin ^{2}{\frac {\Delta \theta }{2}}}{\Delta \theta }}=\lim _{\Delta \theta \to 0}\sin {\frac {\Delta \theta }{2}}\cdot {\frac {\sin {\frac {\Delta \theta }{2}}}{\frac {\Delta \theta }{2}}}=0\cdot 1=0$.	By 39, §1, and §22.

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