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

 Elements of the Differential and Integral Calculus by William Anthony GranvilleChapter V, § 54–61

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

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

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

EXAMPLES

Differentiate the following:

 1. $y = sin ax^2\$. $\frac{dy}{dx}$ $= \cos ax^2 \frac{d}{dx}(ax^2)$ by XI [$v = ax^2$.] 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. $s = \cos 3 ax$. $\frac{ds}{dx}$ $= -3a \sin 3 ax$. 8. $s = \cot(2t^2 + 3)$. $\frac{ds}{dt}$ $= -4 t \csc^2 (2t^2 + 3)$. 9. $f(y) = \sin 2y \cos y$. $f'(y)$ $= 2 \cos 2y \cos y - \sin 2y \sin y$. 10. $F(x) = \cot^2 5x$ $F'(x)$ $= -10 \cot 5x \csc^2 5x$. 11. $F(\theta) = \tan \theta - \theta$. $F'(\theta)$ $= \tan^2 \theta$. 12. $f(\phi) = \phi \sin \phi + \cos \phi$ $f'(\phi)$ $= \phi \cos \phi$. 13. $f(t) = \sin^3 t \cos t$ $f'(t)$ $= \sin^2 t(3\cos^t - \sin^2 t)$. 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. $\frac{d}{d\theta} a(1 - \cos \theta) = a sin \theta$.

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. $y = \log \tan \left ( \frac{\pi}{4} + \frac{x}{2} \right )$. $\frac{dy}{dx}$ $= \frac{1}{\cos x}$. 33. $f(x) = \sin(x + a) \cos(x - a)$ $f'(x)$ $= \cos 2x$. 34. $y = a^{\tan nx}$. $y'$ $= na^{\tan nx} \sec^2 nx \log a$. 35. $y = e^{\cos x} \sin x$. $\ y'$ $= e^{\cos x} (\cos x - \sin^2 x)\$. 36. $y = e^x \log \sin x$. $y'$ $= e^x(\cot x + \log \sin x)$.

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

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 $\ y$ $= \arcsin\ v$;[1] 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 $\ y$ $= \arccos\ v$;[2] 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 $\ y$ $=\ \arctan v$;[3] 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 $\arccot u$.[4]

Following the method of the last section, we get

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

59. Differentiation of $\arcsec u$.

 Let $\ y$ $=\ \arcsec v$;[5] 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 ∴ $\frac{d}{dx} (\arcsec v)$ $= \frac{\frac{dv}{dx}}{v \sqrt{v^2 - 1}}$.

60. Differentiation of $\arccsc v$.[6]

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

61. Differentiation of $\operatorname{arc vers} v$.

 Let $\ y$ $= \operatorname{arcvers} v$;[7] then $\ v$ $= \operatorname{vers} y$. 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 $\frac{dy}{dx}$ $= \frac{1}{\sin y} \cdot \frac{dv}{dx}$ $= \frac{1}{\sqrt{2v - v^2}} \frac{dv}{dx}$ [$\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - (1 - \operatorname{vers} y)^2} = \sqrt{2v - v^2}$, the plus sign of the radical being taken, since $\sin y$ is positive for all values of $y$ between 0 and $\pi$ inclusive.] XXIV ∴ $\frac{d}{dx} (\operatorname{arcvers} v)$ $= \frac{\frac{dv}{dx}}{\sqrt{2v - v^2}}$.

1. It should be remembered that this function is defined only for values of $v$ between -1 and +1 inclusive and that $y$ (the function) is many-valued, there being infinitely many arcs whose sines will equal $v$. Thus, in the figure (the locus of $y = \arcsin v$), when $v = OM, y = MP_1, MP_2, MP_3, \cdots, MQ_1 MQ_2, \cdots$.
In the above discussion, in order to make the function single-valued; only values of $y$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ inclusive (points on arc $QOP$) are considered; that is, the arc of smallest numerical value whose sine is $v$.
2. This function is defined only for values of $v$ between -1 and +1 inclusive, and is many-valued. In the figure (the locus of $y = \arccos v$), when $v=OM, y=MP_1, MP_2, \cdots, MQ_1 MQ_2, \cdots$.
In order to make the function single-valued, only values of $y$ between 0 and π inclusive are considered; that is, $y$ the smallest positive arc whose cosine is $v$. Hence we confine ourselves to arc QP of the graph.
3. This function is defined for all values of $v$, and is many-valued, as is clearly shown by its graph. In order to make it single-valued, only values of $y$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ are considered; that is, the arc of smallest numerical value whose tangent is $v$ (branch $AOE$).
4. This function is defined for all values of $v$, and is many-valued, as is seen from its graph (Fig. a). In order to make it single-valued, only values of $y$ between 0 and $\pi$ are considered; that is, the smallest positive arc whose cotangent is $v$. Hence we confine ourselves to branch AB.
5. This function is defined for all values of $v$ except those lying between -1 and +1, and is seen to be many-valued. To make the function single-valued, $y$ is taken as the arc of smallest numerical value whose secant is $v$. This means that if $v$ is positive, we confine ourselves to points on arc AB (Fig. b), $y$ taking on values between 0 and $\frac{\pi}{2}$ (0 may be included); and if $v$ is negative, we confine ourselves to points on arc DC, $y$ taking on values between $-\pi$ and $-\frac{\pi}{2}$ ($-\pi$ may be included).
6. This function is defined for all values of $v$ except those lying between -1 and +1, and is seen to be many-valued. To make the function single-valued, $y$ is taken as the arc of smallest numerical value whose cosecant is $v$. This means that if $v$ is positive, we confine ourselves to points on the arc AB (Fig. a), $y$ taking on values between 0 and $\frac{\pi}{2}$ ($\frac{\pi}{2}$ may be included); and if $v$ is negative, we confine ourselves to points on the arc CD, $y$ taking on values between $-\pi$ and $-\frac{\pi}{2}$ ($-\frac{\pi}{2}$ may be included).
7. Defined only for values of $v$ between 0 and 2 inclusive, and is many-valued. To make the function continuous, $y$ is taken as the smallest positive arc whose versed sine is $v$; that is, $y$ lies between 0 and $\pi$ inclusive. Hence we confine ourselves to arc $OP$ of the graph (Fig. a).