Elements of the Differential and Integral Calculus/Chapter V part 3

54. Differentiation of $\mathrm{vers}v$.

Let	$y$	$=\mathrm{vers}v$.
By Trigonometry this may be written
	$y$	$=1-\cos v$.
Differentiating,
	$\frac{dy}{dx}$	$=\sin v\frac{dv}{dx}$.
XVII	∴ $\frac{d}{dx}(\mathrm{vers}v)$	$=\sin v\frac{dv}{dx}$.

In the derivation of our formulas so far it has been necessary to apply the General Rule, p. 29 [§ 31] (i.e. the four steps), only for the following:
III	$\frac{d}{dx}(u+v-w)$	$=\frac{du}{dx}+\frac{dv}{dx}-\frac{dw}{dx}$	Algebraic sum.
V	$\frac{d}{dx}(uv)$	$=u\frac{dv}{dx}+v\frac{du}{dx}$.	Product.
VII	$\frac{d}{dx}\left(\frac{u}{v}\right)$	$=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$.	Quotient.
VIII	$\frac{d}{dx}({\log }_{a}v)$	$={\log }_{a}e\frac{\frac{dv}{dx}}{v}$.	Logarithm.
XI	$\frac{d}{dx}(\sin v)$	$=\cos v\frac{dv}{dx}$	Sine.
XXV	$\frac{dy}{dx}$	$=\frac{dy}{dv}\cdot \frac{dv}{dx}$.	Function of a function.
XXVI	$\frac{dy}{dx}$	$=\frac{1}{\frac{dx}{dy}}$.	Inverse functions.

Not only do all the other formulas we have deduced depend on these, but all we shall deduce hereafter depend on them as well. Hence it follows that the derivation of the fundamental formulas for differentiation involves the calculation of only two limits of any difficulty, viz.,

	$\underset{v\to 0}{\lim }\frac{\sin v}{1}$	$=1$	by § 22, p. 21
and	$\underset{v\to 0}{\lim }(1+v{)}^{\frac{1}{v}}$	$=e$.	By § 23, p. 22

EXAMPLES

Differentiate the following:

1. $y=sina{x}^2$.
$\frac{dy}{dx}$	$=\cos a{x}^2\frac{d}{dx}(a{x}^2)$	by XI
	[$v=a{x}^2$.]
2. $y=\tan \sqrt{1-x}$.
$\frac{dy}{dx}$	$={\sec }^2\sqrt{1-x}\frac{d}{dx}(1-x{)}^{\frac{1}{2}}$	by XIII
	[$v=\sqrt{1-x}$.]
	$={\sec }^2\sqrt{1-x}\cdot \frac{1}{2}(1-x{)}^{-\frac{1}{2}}(-1)$.
	$=-\frac{{\sec }^2\sqrt{1-x}}{2\sqrt{1-x}}$.
3. $y={\cos }^{3}x$.
This may also be written,
$y$	$=(\cos x{)}^3$.
$\frac{dy}{dx}$	$=3(\cos x{)}^2\frac{d}{dx}(\cos x)$	by VI
	[$v=\cos x$ and $n=3$.]
	$=3{\cos }^{2}x(-\sin x)$	by XII
	$=-3\sin x{\cos }^{2}x$
4. $y=\sin nx{\sin }^{n}x$.
$\frac{dy}{dx}$	$=\sin nx\frac{d}{dx}(\sin x{)}^n+{\sin }^{n}x\frac{d}{dx}(\sin nx)$	by V
	[$v=\sin nx$ and $v={\sin }^{n}x$.]
	$=\sin nx\cdot n(\sin x{)}^{n-1}\frac{d}{dx}(\sin x)+{\sin }^{n}x\cos nx\frac{d}{dx}(nx)$	by VI and XI
	$=n\sin nx\cdot {\sin }^{n-1}x\cos x+n{\sin }^{n}x\cos nx$
	$=n{\sin }^{n-1}x(\sin nx\cos x+\cos nx\sin x)$
	$=n{\sin }^{n-1}x\sin (n+1)x$.

5. $y=\sec ax$.	Ans. $\frac{dy}{dx}$	$a\sec ax\tan ax$.
6. $y=\tan (ax+b)$.	$\frac{dy}{dx}$	$=a{\sec }^2(ax+b)$.
7. $s=\cos 3ax$.	$\frac{ds}{dx}$	$=-3a\sin 3ax$.
8. $s=\cot (2t^2+3)$.	$\frac{ds}{dt}$	$=-4t{\csc }^2(2t^2+3)$.
9. $f(y)=\sin 2y\cos y$.	$f^{\prime }(y)$	$=2\cos 2y\cos y-\sin 2y\sin y$.
10. $F(x)={\cot }^{2}5x$	$F^{\prime }(x)$	$=-10\cot 5x{\csc }^{2}5x$.
11. $F(\theta )=\tan \theta -\theta$.	$F^{\prime }(\theta )$	$={\tan }^2\theta$.
12. $f(\varphi )=\varphi \sin \varphi +\cos \varphi$	$f^{\prime }(\varphi )$	$=\varphi \cos \varphi$.
13. $f(t)={\sin }^{3}t\cos t$	$f^{\prime }(t)$	$={\sin }^{2}t(3{\cos }^t-{\sin }^{2}t)$.
14. $r=a\cos 2\theta$.	$\frac{dr}{d\theta }$	$=-2a\sin 2\theta$.

15. $\frac{d}{dx}{\sin }^{2}x=\sin 2x$.

16. $\frac{d}{dx}{\cos }^3{x}^2=-6x{\cos }^2{x}^2\sin x^2$.

17. $\frac{d}{dt}\csc \frac{t^2}{2}=-t\csc \frac{t^2}{2}\cot \frac{t^2}{2}$.

18. $\frac{d}{ds}a\sqrt{\cos 2s}=-\frac{a\sin 2s}{\sqrt{\cos 2s}}$.

19. $\frac{d}{d\theta }a(1-\cos \theta )=asin\theta$.

20. $\frac{d}{dx}(\log \cos x)=-\tan x$.

21. $\frac{d}{dx}(\log \tan x)=\frac{2}{\sin 2x}$.

22. $\frac{d}{dx}(\log {\sin }^{2}x)=2\cot x$.

23. $\frac{d}{dt}\cos \frac{a}{t}=\frac{a}{t^2}\sin \frac{a}{t}$.

24. $\frac{d}{d\theta }\sin \frac{1}{{\theta }^2}=-\frac{2}{{\theta }^3}\cos \frac{1}{{\theta }^2}$.

25. $\frac{d}{dx}{e}^{\sin x}=e^{\sin x}\cos x$.

26. $\frac{d}{dx}\sin (\log x)=\frac{\cos (\log x)}{x}$.

27. $\frac{d}{dx}\tan (\log x)=\frac{{\sec }^2(\log x)}{x}$.

28. $\frac{d}{dx}a{\sin }^3\frac{\theta }{3}=a{\sin }^2\frac{\theta }{3}\cos \frac{\theta }{3}$.

29. $\frac{d}{d\alpha }\sin (\cos \alpha )=-\sin \alpha \cos (\cos \alpha )$.

30. $\frac{d}{dx}\frac{\tan x-1}{\sec x}=\sin x+\cos x$.

31. $y=\log \sqrt{\frac{1+\sin x}{1-\sin x}}$.	$\frac{dy}{dx}$	$=\frac{1}{\cos x}$.
32. $y=\log \tan (\frac{\pi }{4}+\frac{x}{2})$.	$\frac{dy}{dx}$	$=\frac{1}{\cos x}$.
33. $f(x)=\sin (x+a)\cos (x-a)$	$f^{\prime }(x)$	$=\cos 2x$.
34. $y=a^{\tan nx}$.	$y^{\prime }$	$=n{a}^{\tan nx}{\sec }^{2}nx\log a$.
35. $y=e^{\cos x}\sin x$.	$y^{\prime }$	$=e^{\cos x}(\cos x-{\sin }^{2}x)$.
36. $y=e^x\log \sin x$.	$y^{\prime }$	$=e^x(\cot x+\log \sin x)$.

37. Differentiate the following functions:

(a) $\frac{d}{dx}\sin 5x^2$.	(f) $\frac{d}{dx}\csc (\log x)$.	(k) $\frac{d}{dt}{e}^{a-b\cos t}$.
(b) $\frac{d}{dx}\cos (a-bx)$.	(g) $\frac{d}{dx}{\sin }^{3}2x$	(l) $\frac{d}{dt}\sin \frac{t}{3}{\cos }^2\frac{t}{3}$.
(c) $\frac{d}{dx}\tan \frac{ax}{b}$.	(h) $\frac{d}{dx}{\cos }^2(\log x)$.	(m) $\frac{d}{d\theta }\cot \frac{b}{{\theta }^2}$.
(d) $\frac{d}{dx}\cot \sqrt{ax}$.	(i) $\frac{d}{dx}{\tan }^2\sqrt{1-x^2}$.	(n) $\frac{d}{d\varphi }\sqrt{1+{\cos }^2\varphi }$.
(e) $\frac{d}{dx}\sec e^{3x}$.	(j) $\frac{d}{dx}\log ({\sin }^{2}ax)$.	(o) $\frac{d}{ds}\log \sqrt{1-2{\sin }^{2}s}$.

38. $\frac{d}{dx}(x^n{e}^{\sin x})=x^{n-1}{e}^{\sin x}(n+x\cos x)$.

39. $\frac{d}{dx}(e^{ax}\cos mx)=e^{ax}(a\cos mx-m\sin mx)$.

40. $f(\theta )=\frac{1+\cos \theta }{1-\cos \theta }$.	$f^{\prime }(\theta )$	$=-\frac{2\sin \theta }{(1-\cos \theta {)}^2}$.
41. $f(\varphi )=\frac{e^{a\varphi }(a\sin \varphi -\cos \varphi )}{a^2+1}$.	$f^{\prime }(\varphi )$	$=e^{a\varphi }\sin \varphi$.
42. $f(s)=(s\cot s{)}^2$.	$f^{\prime }(s)$	$=2s\cot s(\cot s-s{\csc }^{2}s)$.
43. $r=\frac{1}{3}{\tan }^3\theta -\tan \theta +\theta$.	$\frac{dr}{d\theta }$	$={\tan }^4\theta$.
44. $y=x^{\sin x}$.	$\frac{dy}{dx}$	$=x^{\sin x}(\frac{\sin x}{x}+\log x\cos x)$.
45. $y=(\sin x{)}^x$.	$y^{\prime }$	$=(\sin x{)}^x[\log \sin x+x\cot x]$.
46. $y=(\sin x{)}^{\tan x}$.	$y^{\prime }$	$=(\sin x{)}^{\tan x}(1+{\sec }^{2}x\log \sin x)$.

47. Prove $\frac{d}{dx}\cos v=-\sin v\frac{dv}{dx}$, using the General Rule.

48. Prove $\frac{d}{dx}\cot v=-{\csc }^{2}v\frac{dv}{dx}$ by replacing $\cot v$ by $\frac{\cos c}{\sin v}$.

55. Differentiation of $\arcsin v$.

Let	$y$	$=\arcsin v$;
then	$v$	$=\sin y$.
Differentiating with respect to $y$ by XI,
	$\frac{dv}{dy}$	$=\cos y$;
therefore	$\frac{dy}{dv}$	$=\frac{1}{\cos y}$.	By (C), p. 46 [§ 43]
But since $v$ is a function of $x$, this may be substituted in
	$\frac{dy}{dx}$	$=\frac{dy}{dv}\cdot \frac{dv}{dx}$	(A), p. 45 [§ 42]
giving	$\frac{dy}{dx}$	$=\frac{1}{\cos y}\cdot \frac{dv}{dx}$.
		$=\frac{1}{\sqrt{1-v^2}}\frac{dv}{dx}$.
$[\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-v^2}$, the positive sign of the radical being taken, since $\cos y$ is positive for all values of $y$ between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$ inclusive.]
XVIII	∴ $\frac{d}{dx}(\arcsin v)$	$=\frac{\frac{dv}{dx}}{\sqrt{1-v^2}}$.

56. Differentiation of $\arccos v$.

Let	$y$	$=\arccos v$;
then	$y$	$=\cos y$.
Differentiating with respect to $y$ by XII,
	$\frac{dv}{dy}$	$=-\sin y$.
therefore	$\frac{dy}{dv}$	$=-\frac{1}{\sin y}$.	By (C), p. 46 [§ 43]
But since $v$ is a function of $x$, this may be substituted in the formula
	$\frac{dy}{dx}$	$=\frac{dy}{dv}\cdot \frac{dv}{dx}$,	(A), p. 45 [§ 42]
giving	$\frac{dy}{dx}$	$=-\frac{1}{\sin y}\cdot \frac{dv}{dx}$
		$=-\frac{1}{\sqrt{1-v^2}}\frac{dv}{dx}$.
[ $\sin y=\sqrt{1-{\cos }^{2}y}=\sqrt{1-v^2}$, the plus sign of the radical being taken, since $\sin y$ is positive for all values of y between 0 and π inclusive.]
XIX	∴ $\frac{d}{dx}(\arccos v)$	$=-\frac{\frac{dv}{dx}}{\sqrt{1-v^2}}$.

57. Differentiation of $\arctan v$.

Let	$y$	$=\arctan v$;
then	$y$	$=\tan y$.
Differentiating with respect to $y$ by XIV,
	$\frac{dv}{dy}$	$={\sec }^{2}y$;
therefore	$\frac{dy}{dv}$	$=\frac{1}{{\sec }^{2}y}$.	By (C), p. 46 [§ 43]
But since $v$ is a function of $x$, this may be substituted in the formula
	$\frac{dy}{dx}$	$=\frac{dy}{dv}\cdot \frac{dv}{dx}$,	(A), p. 45 [§ 42]
giving	$\frac{dy}{dx}$	$=\frac{1}{{\sec }^{2}y}\cdot \frac{dv}{dx}$
		$=\frac{1}{1+v^2}\frac{dv}{dx}$.
	[${\sec }^{2}y=1+{\tan }^{2}y=1+v^2$]
XX	∴ $\frac{d}{dx}(\arctan v)$	$=\frac{\frac{dv}{dx}}{1+v^2}$

58. Differentiation of $\mathrm{arccot}u$.

Following the method of the last section, we get

XXI $\frac{d}{dx}(\mathrm{arccot}v)=-\frac{\frac{dv}{dx}}{1+v^2}$.

59. Differentiation of $\mathrm{arcsec}u$.

Let	$y$	$=\mathrm{arcsec}v$;
then	$v$	$=\sec y$.
Differentiating with respect to $y$ by IV,
	$\frac{dv}{dy}$	$=\sec y\tan y$;
therefore	$\frac{dy}{dv}$	$=\frac{1}{\sec y\tan y}$	By (C), p. 46 [§ 43]
But since $v$ is a function of $x$, this may be substituted in the formula
	$\frac{dy}{dx}$	$=\frac{dy}{dv}\cdot \frac{dv}{dx}$,	(A), p. 45 [§ 42]
giving	$\frac{dy}{dx}$	$=\frac{1}{\sec y\tan y}\frac{dv}{dx}$
		$=\frac{1}{v\sqrt{v^2-1}}\frac{dv}{dx}$.
[$\sec y=v$, and $\tan y=\sqrt{\sec y-1}=\sqrt{v^2-1}$, the plus sign of the radical being taken, since $\tan y$ is positive for an values of $y$ between 0 and $\frac{\pi }{2}$ and between $-\pi$ and $-\frac{\pi }{2}$, including 0 and $-\pi$].
XXII	∴ $\frac{d}{dx}(\mathrm{arcsec}v)$	$=\frac{\frac{dv}{dx}}{v\sqrt{v^2-1}}$.

60. Differentiation of $\mathrm{arccsc}v$.

Let	$y$	$=\mathrm{arccsc}v$;
then	$v$	$=\csc y$.
Differentiating with respect to $y$ by XVI and following the method of the last section, we get
XXIII	$\frac{d}{dx}(\mathrm{arccsc}v)$	$=-\frac{\frac{dv}{dx}}{v\sqrt{v^2-1}}$.

61. Differentiation of $\mathrm{arcvers}v$.

Let	$y$	$=\mathrm{arcvers}v$;
then	$v$	$=\mathrm{vers}y$.
Differentiating with respect to $y$ by XVII,
	$\frac{dv}{dy}$	$=\sin y$;
therefore	$\frac{dy}{dv}$	$=\frac{1}{\sin y}$	By (C), p. 46 [§ 43]
But since $v$ is a function of $x$, this may be substituted in the formula
	$\frac{dy}{dx}$	$=\frac{dy}{dv}\cdot \frac{dv}{dx}$	(A), p. 45 [§ 42]
giving	$\frac{dy}{dx}$	$=\frac{1}{\sin y}\cdot \frac{dv}{dx}$
		$=\frac{1}{\sqrt{2v-v^2}}\frac{dv}{dx}$
[$\sin y=\sqrt{1-{\cos }^{2}y}=\sqrt{1-(1-\mathrm{vers}y{)}^2}=\sqrt{2v-v^2}$, the plus sign of the radical being taken, since $\sin y$ is positive for all values of $y$ between 0 and $\pi$ inclusive.]
XXIV	∴ $\frac{d}{dx}(\mathrm{arcvers}v)$	$=\frac{\frac{dv}{dx}}{\sqrt{2v-v^2}}$.