# Elements of the Differential and Integral Calculus/Chapter V part 4

 Elements of the Differential and Integral Calculus by William Anthony Granville Chapter V, § 62–63
EXAMPLES

Differentiate the following:

 1. $y = \arctan ax^2$. Solution. $\frac{dy}{dx}$ $= \frac{\frac{d}{dx} (ax^2)}{1 + (ax^2)^2}$ by XX [$v = ax^2$.] $= \frac{2ax}{1 + a^2x^4}$. 2. $y = \arcsin (3x - 4x^3)$. Solution. $\frac{dy}{dx}$ $\frac{\frac{d}{dx}(3x - 4x^3}{\sqrt{1 - (3x - 4x^3)^2}}$ by XVIII [$v = 3x - 4x^3$.] $\frac{3 - 12x^2}{\sqrt{1 - 9x^2 + 24x^4 - 16x^6}} = \frac{3}{\sqrt{1 - x^2}}$. 3. $\arcsec \frac{x^2 + 1}{x^2 - 1}$ Solution. $\frac{dy}{dx}$ $= \frac{\frac{d}{dx} \left ( \frac{x^2 + 1}{x^2 - 1} \right )}{\frac{x^2 + 1}{x^2 - 1} \sqrt{\left ( \frac{x^2 + 1}{x^2 - 1} \right )^2 - 1}}$ by XXII $v = \left [ \frac{x^2 + 1}{x^2 - 1} \right ]$. $= \frac{\frac{(x^2 - 1)2x - (x^2 + 1)2x}{(x^2 - 1)^2}}{\frac{x^2 + 1}{x^2 - 1} \cdot \frac{2x}{x^2 - 1}} = -\frac{2}{x^2 + 1}$

4. $\frac{d}{dx} \arcsin \frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}}$.

5. $\frac{d}{dx} \arccot (x^2 - 5) = \frac{-2x}{1 + (x^2 - 5)^2}$.

6. $\frac{d}{dx} \arctan \frac{2x}{1 - x^2} = \frac{2}{1 + x^2}$.

7. $\frac{d}{dx} \arccsc \frac{1}{2x^2 - 1} = \frac{2}{\sqrt{1 - x^2}}$.

8. $\frac{d}{dx} \operatorname{arcvers} 2x^2 = \frac{2}{\sqrt{1 - x^2}}$.

9. $\frac{d}{dx} \arctan \sqrt{1 - x} = -\frac{1}{2 \sqrt{1 - x} (2 - x)}$.

10. $\frac{d}{dx} \arccsc \frac{3}{2x} = \frac{2}{9 - 4x^2}$.

11. $\frac{d}{dx} \operatorname{arcvers} \frac{2x^2}{1 + x^2} = \frac{2}{1 + x^2}$.

12. $\frac{d}{dx} \arctan \frac{x}{a} = \frac{a}{a^2 + x^2}$.

13. $\frac{d}{dx} \arcsin \frac{x + 1}{\sqrt{2}} = \frac{1}{\sqrt{1 - 2x - x^2}}$.

 14. $f(x) = x\sqrt{a^2 - x2} + a^2 \arcsin \frac{x}{a}$ $\ f'(x)$ $=\ 2\sqrt{a^2 - x^2}$. 15. $f(x) = \sqrt{a^2 - x^2} + a \arcsin \frac{x}{a}$ $\ f'(x)$ $= \left ( \frac{a - x}{a + x} \right )^{\frac{1}{2}}$ 16. $x = r \operatorname{arcvers} \frac{y}{r} - \sqrt{2ry - y^2}$. $\frac{dx}{dy}$ $= \frac{y}{\sqrt{2ry - y^2}}$. 17. $\theta = \arcsin (3r - 1)\$ $\frac{d\theta}{dr}$ $= \frac{3}{\sqrt{6r - 9r^2}}$. 18. $\phi = \arctan \frac{r + a}{1 - ar}$ $\frac{d\phi}{dr}$ $= \frac{1}{1 + r^2}$. 19. $s = \arcsec \frac{1}{\sqrt{1 - t^2}}$. $\frac{ds}{dt}$ $\frac{1}{\sqrt{1 - t^2}}$.

20. $\frac{d}{dx} ( x \arcsin x) = \arcsin x + \frac{x}{\sqrt{1 - x^2}}$.

21. $\frac{d}{d\theta} (\tan \theta \arctan \theta) = \sec^2 \theta \arctan \theta \frac{\tan \theta}{1 + \theta^2}$.

22. $\frac{d}{dt}[\log (\arccos t)] = -\frac{1}{\arccos t \sqrt{1 - t^2}}$.

 23. $f(y) = \arccos (\log y)\$. $\ f'(y)$ $= -\frac{1}{y\sqrt{1 - (\log y)^2}}$. 24. $f(\theta) = \arcsin \sqrt{\sin \theta}$ $\ f'(\theta)$ $= \frac{1}{2} \sqrt{1 + \csc \theta}$. 25. $f(\phi) = \arctan \sqrt{\frac{1 - \cos \phi}{1 + \cos \phi}}$. $\ f'(\phi)$ $= \frac{1}{2}$ 26. $p = e^{\arctan q}\$. $\frac{dp}{dq}$ $= \frac{e^{\arctan q}}{1 + q^2}$. 27. $u = \arctan \frac{e^v - e^{-v}}{2}$ $\frac{du}{dv}$ $= \frac{2}{e^v + e^{-v}}$ 28. $s = \arccos \frac{e^t - e^{-t}}{e^t + e^{-t}}$. $\frac{ds}{dt}$ $= -\frac{2}{e^v + e^{-v}}$. 29. $y = x^{\arcsin x}\$. $\ y'$ $= x^{\arcsin x} \left ( \frac{\arcsin x}{x} + \frac{\log x}{\sqrt{1 - x^2}} \right )$. 30. $y = e^{x^x} \arctan x$ $\ y'$ $= e^{x^x} \left [ \frac{1}{1 + x^2} + x^x \arctan x (1 + \log x) \right ]$ 31. $y = \arcsin( \sin x )\$. $\ y'$ $=\ 1$. 32. $y = \arctan \frac{4 \sin x}{3 + 5 \cos x}$. $\ y'$ $= \frac{4}{5 + 3 \cos x}$. 33. $y = \arccot \frac{a}{x} + \log \sqrt{\frac{x - a}{x + a}}$. $\ y'$ $= \frac{2ax^2}{x^4 - a^4}$. 34. $y = \log \left ( \frac{1 + x}{1 - x} \right )^{\frac{1}{4}} - \frac{1}{2} \arctan x$. $\ y'$ $= \frac{x^2}{1 - x^4}$. 35. $y = \sqrt{1 - x^2} \arcsin x - x$. $\ y'$ $= - \frac{x \arcsin x}{\sqrt{1 - x^2}}$.

36. Differentiate the following functions:

 (a) $\frac{d}{dx} \arcsin 2x^2$ (f) $\frac{d}{dt} t^3 \arcsin \frac{t}{3}$. (k) $\frac{d}{dy} \arcsin \sqrt{1 - y^2}$. (b) $\frac{d}{dx} \arctan a^2x$. (g) $\frac{d}{dt} e^{\arctan at}$. (l) $\frac{d}{dz} \arctan ( \log 3 az )$. (c) $\frac{d}{dx} \arcsec \frac{x}{a}$. (h) $\frac{d}{d\phi} \tan \phi^2 \cdot \arctan \phi^{\frac{1}{2}}$. (m) $\frac{d}{ds}(a^2 + s^2)\arcsec \frac{s}{2}$. (d) $\frac{d}{dx} x \arccos x$. (i) $\frac{d}{d\theta} \arcsin a^{\theta}$. (n) $\frac{d}{d\alpha} \arccot \frac{2\alpha}{3}$. (e) $\frac{d}{dx} x^2 \arccot ax$. (j) $\frac{d}{d\theta} \arctan \sqrt{1 + \theta^2}$. (o) $\frac{d}{dt} \sqrt{1 - t^2} \arcsin t$.

Formulas (A), p. 45 [§ 42], for differentiating a function of a function, and (C), p. 46 [§ 43], for differentiating inverse junctions, have been added to the list of formulas at the beginning of this chapter as XXV and XXVI respectively.

In the next eight examples, first find $\frac{dy}{dv}$ and $\frac{dv}{dx}$ by differentiation and then substitute the results in

 $\frac{dy}{dx}$ $= \frac{dy}{dv} \cdot \frac{dv}{dx}$ by XXV

to find $\frac{dy}{dx}$.[1]

In general our results should be expressed explicitly in terms of the independent variable; that is, $\frac{dy}{dx}$ in terms of $x, \frac{dx}{dy}$ in terms of $y, \frac{d\phi}{d\theta}$ in terms of $\theta$, etc.

37. $y = 2v^2 - 4, v = 3x^2 + 1$.

$\frac{dy}{dv} = 4v; \frac{dv}{dx} = 6x$; substituting in XXV,

$\frac{dy}{dx} = 4v \cdot 6x = 24x(3x^2 + 1)$.

38. $y = \tan 2v, v = \arctan(2x - 1)$.

$\frac{dy}{dv} = 2 \sec^2 2v; \frac{dv}{dx} = \frac{1}{2x^2 - 2x + 1}$; substituting in XXV,

$\frac{dy}{dx} = \frac{2 \sec^2 2v}{2x^2 - 2x + 1} = 2 \frac{\tan^2 2v + 1}{2x^2 - 2x + 1} = \frac{2x^2 - 2x + 1}{2(x - x^2)^2}$

[Since $v = \arctan (2x - 1), \tan v = 2x - 1, \tan 2v = \frac{2x - 1}{2x - 2x^2}$.]

 39. $y = 3 v^2 - 4v + 5, v = 2x^3 - 5\$. $\frac{dy}{dx}$ $=\ 72x^5 - 204x^2$. 40. $y = \frac{2v}{3v - 2}, v = \frac{x}{2x - 1}$. $\frac{dy}{dx}$ $= \frac{4}{(x - 2)^2}$ 41. $y = \log(a^2 - v^2)\$. $\frac{dy}{dx}$ $=\ -2 \tan x$. 42. $y = \arctan (a + v), v = e^x\$ $\frac{dy}{dx}$ $=\ \frac{e^x}{1 + (a + e^x)^2}$. 43. $r = e^{2s} + e^s, s = \log(t - t^2)\$. $\frac{dr}{dt}$ $=\ 4t^3 - 6t^2 + 1$.

In the following examples first find $\frac{dx}{dy}$ by differentiation and then substitute in

 $\frac{dy}{dx}$ $=\ \frac{1}{\frac{dx}{dy}}$ by XXVI

to find $\frac{dy}{dx}$.

 44. $x = y\sqrt{1 + y}$. $\frac{dy}{dx}$ $=\ \frac{2\sqrt{1 + y}}{2 + 3y} = \frac{2x}{2y + 3y^2}$. 45. $x = \sqrt{1 + \cos y}$. $\frac{dy}{dx}$ $=\ -\frac{2 \sqrt{1 + \cos y}}{\sin y} = -\frac{2}{\sqrt{2 - x^2}}$. 46. $x = \frac{y}{1 + \log y}$. $\frac{dy}{dx}$ $=\ \frac{(1 + \log y)^2}{\log y}$. 47. $x = a \log \frac{a + \sqrt{a^2 - y^2}}{y}$. $\frac{dy}{dx}$ $=\ -\frac{y \sqrt{a^2 - y^2}}{a^2}$. 48. $x = r \operatorname{arcvers} \frac{y}{r} - \sqrt{2ry - y^2}$. $\frac{dy}{dx}$ $=\ \sqrt{\frac{2r - y}{y}}$.

49. Show that the geometrical significance of XXVI is that the tangent makes complementary angles with the two coördinate axes.

62. Implicit functions. When a relation between $x$ and $y$ is given by means of an equation not solved for $y$, then $y$ is called an implicit function of $x$. For example, the equation

$x^2 - 4y = 0$

defines $y$ as an implicit function of $x$. Evidently $x$ is also defined by means of this equation as an implicit function of $y$. Similarly,

$x^2 + y^2 + z^2 - a^2 = 0$

defines anyone of the three variables as an implicit function of the other two.

It is sometimes possible to solve the equation defining an implicit function for one of the variables and thus change it into an explicit function. For instance, the above two implicit functions may be solved for $y$, giving

 $\ y$ $= \frac{x^2}{4}$ and $\ y$ $= \pm \sqrt{a^2 - x^2 - z^2}$;

the first showing $y$ as an explicit function of $x$, and the second as an explicit function of $x$ and $z$. In a given case, however, such a solution may be either impossible or too complicated for convenient use.

The two implicit functions used in this article for illustration may be respectively denoted by

 $f(x, y)$ $=0$ and $F(x, y, z)$ $= 0$;

63. Differentiation of implicit functions. When $y$ is defined as an implicit function of $x$ by means of an equation in the form

(A) $f(x, y)= 0$,

it was explained in the last section how it might be inconvenient to solve for $y$ in terms of $x$; that is, to find $y$ as an explicit function of $x$ so that the formulas we have deduced in this chapter may be applied directly. Such, for instance, would be the case for the equation

(B) $ax^6 + 2x^3y - y^7x - 10 = 0$.

We then follow the rule:

Differentiate, regarding $y$ as a function of $x$, and put the result equal to zero.[2] That is,

(C) $\frac{d}{dx} f(x, y) = 0$.

Let us apply this rule in finding $\frac{dy}{dx}$ from (B).

 $\frac{d}{dx}(ax^6 + 2x^3y - y^7x - 10) = 0$; by (C) $\frac{d}{dx}(ax^6) + \frac{d}{dx}(2x^3y) - \frac{d}{dx}(y^7x) - \frac{d}{dx}(10) = 0$; $6ax^5 + 2x^3 \frac{dy}{dx} + 6x^2y - y^7 - 7xy^6\frac{dy}{dx} = 0$; $(2x^3 - 7xy^6)\frac{dy}{dx} = y^7 - 6ax^5 - 6x^2y$; $\frac{dy}{dx} = \frac{y^7 - 6ax^5 - 6x^2y}{2x^3 - 7xy^6}$ Ans.

The student should observe that in general the result will contain both $x$ and $y$.

EXAMPLES

Differentiate the following by the above rule:

 1. $y^2 = 4px\$. $\frac{dy}{dx}$ $=\ \frac{2p}{y}$. 2. $x^2 + y^2 = r^2\$. $\frac{dy}{dx}$ $=\ -\frac{x}{y}$. 3. $b^2x^2 + a^2y^2 = a^2b^2\$. $\frac{dy}{dx}$ $=\ -\frac{b^2x}{a^2y}$. 4. $y^3 - 3y + 2ax = 0\$. $\frac{dy}{dx}$ $=\ \frac{2a}{3(1 - y^2)}$. 5. $x^{\frac{1}{2}} + y^{\frac{1}{2}} = a^{\frac{1}{2}}\$. $\frac{dy}{dx}$ $=\ -\sqrt{\frac{y}{x}}$. 6. $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$. $\frac{dy}{dx}$ $=\ -\sqrt[3]{\frac{y}{x}}$. 7. $\left ( \frac{x}{a} \right )^2 + \left ( \frac{y}{b} \right )^{\frac{2}{3}} = 1\$. $\frac{dy}{dx}$ $=\ -\frac{3b^{\frac{2}{3}}xy^{\frac{1}{3}}}{a^2}$. 8. $y^2 - 2xy + b^2 = 0\$. $\frac{dy}{dx}$ $=\ \frac{y}{y - x}$ 9. $x^3 + y^3 - 3axy = 0\$. $\frac{dy}{dx}$ $=\ \frac{ay - x^2}{y^2 - ax}$. 10. $x^y = y^x\$. $\frac{dy}{dx}$ $=\ \frac{y^2 - xy\log y}{x^2 - xy\log x}$. 11. $\rho^2 = a^2 \cos 2\theta\$. $\frac{d\rho}{d\theta}$ $=\ -\frac{a^2 \sin 2\theta}{\rho}$. 12. $\rho^2 \cos \theta = a^2 \sin 3\theta\$. $\frac{d\rho}{d\theta}$ $=\ \frac{3a^2 \cos 3\theta + \rho^2 \sin \theta}{2\rho \cos \theta}$. 13. $\cos(uv) = cv$. $\frac{du}{dv}$ $=\ \frac{c + u\sin(uv)}{-v\sin(uv)}$ 14. $\theta = \cos(\theta + \phi)\$. $\frac{d\theta}{d\phi}$ $=\ -\frac{\sin(\theta + \phi)}{1 + \sin(\theta + \phi)}$.

15. Find $\frac{dy}{dx}$ from the following equations:

 (a) $x^2 = ay$. (f) $xy + y^2 + 4x = 0$. (k) $\tan x + y^3 = 0$. (b) $x^2 + 4y^2 = 16$. (g) $yx^2 - y^3 = 5$. (l) $\cos y + 3x^2 = 0$. (c) $b^2x^2 - a^2y^2 = a^2b^2$. (h) $x^2 - 2x^3 - y^3$. (m) $x \cot y + y = 0$. (d) $y^2 = x^3 + a$. (i) $x^2y^3 + 4y = 0$. (n) $y^2 = \log x$. (e) $x^2 - y^2 = 16$. (j) $y^2 = \sin 2x$. (o) $e^{x^2} + 2y^3 = 0$.

16. A race track has the form of the circle $x^2 + y^2 = 2500$. The directions OX and OY are east and north respectively, and the unit is 1 rod. If a runner starts east at the extreme north point, in what direction will he be going

 (a) when $25\sqrt{2}$ rods east of OY? Ans. Southeast or southwest. (b) when $25\sqrt{2}$ rods north of OX? Southeast or northeast. (c) when 30 rods west of OY? E. 36° 52′ 12&Prime N. or W. 36° 52′ 12″ N. (d) when 40 rods south of OX? (e) when 10 rods east of OY?

17. An automobile course is elliptic in form, the major axis being 6 miles long and running east and west, while the minor axis is 2 miles long. If a car starts north at the extreme east point of the course, in what direction will the car be going

 (a) when 2 miles west of the starting point? (b) when ½ mile north of the starting point?
MISCELLANEOUS EXAMPLES

Differentiate the following functions:

 1. $\arcsin \sqrt{1 - 4x^2}\$. $\frac{-2}{\sqrt{1 - 4x^2}}$ Ans. 2. $xe^{x^2}$. $e^{x^2}(2x^2 + 1)$. 3. $\log \sin \frac{v}{2}$. $\frac{1}{2} \cot \frac{v}{2}$. 4. $\arccos \frac{a}{y}$. $\frac{a}{y\sqrt{y^2 - a^2}}$. 5. $\frac{x}{\sqrt{a^2 - x^2}}$. $\frac{a^2}{(a^2 - x^2)^{\frac{3}{2}}}$. 6. $\frac{x}{1 + \log x}$. $\frac{log x}{(1 + \log x)^2}$ 7. $\log \sec(1 - 2x)$. $-2\tan(1 - 2x)$. 8. $x^2e^{2 - 3x}$. $xe^{2 - 3x}(2 - 3x)$. 9. $\log \sqrt{\frac{1 - \cos t}{1 + \cos t}}$ $\csc t$ 10. $\arcsin \sqrt{\frac{1}{2}(1 - \cos x)}$. $\frac{1}{2}$ 11. $\arctan \frac{2s}{\sqrt{s^2 - 1}}$. $\frac{2}{(1 - 5s^2)\sqrt{s^2 - 1}}$. 12. $(2x - 1)\sqrt[3]{\frac{2}{1 + x}}$. $\frac{7 + 4x}{3(1 + x)}\sqrt[3]{\frac{2}{1 + x}}$. 13. $\frac{x^3 \arcsin x}{3} + \frac{(x^2 + 2)\sqrt{1 - x^2}}{9}$. $x^2 \arcsin x$.

14. $\tan^2 \frac{\theta}{3} + \log \sec^2 \frac{\theta}{3}$.

15. $\arctan \frac{1}{2}(e^{2x} + e^{-2x})$.

16. $\left ( \frac{3}{x} \right )^{2x}$.

17. $x^{\tan x}$.

18. $\frac{(x + 2)^{\frac{1}{3}} (x^2 - 1)^{\frac{2}{5}}}{x^{\frac{3}{2}}}$.

19. $e^{\sec(1 - 3x)}$.

20. $\arctan \sqrt{1 - x^2}$.

21. $\frac{z^2}{\cos z}$.

22. $e^{\tan x^2}$.

23. $\log \sin^2 \frac{1}{2} \theta$.

24. $e^{ax} \log \sin ax$.

25. $\sin^3 \phi \cos \phi$.

26. $\frac{a}{2\sqrt{(b - cx^n)^m}}$.

27. $\frac{m + x}{1 + m^2} \cdot \frac{e^{m \arctan x}}{\sqrt{1 + x^2}}$.

28. $\tan^2 x - \log \sec^2 x$.

29. $\frac{3 \log (2 \cos x + 3 \sin x) + 2x}{13}$.

30. $\arccot \frac{a}{x} + \log \sqrt{\frac{x - a}{x + a}}$.

31. $(\log \tan 3 - x^2)^3$.

32. $\frac{2 - 3t^{\frac{1}{2}} + 4t^{\frac{1}{3}} + t^2}{t}$.

33. $\frac{(1 + x)(1 - 2x)(2 + x)}{(3 + x)(2 - 3x)}$.

34. $\arctan (\log 3x)$.

35. $\sqrt[3]{(b - ax^m)^n}$.

36. $\log \sqrt{(a^2 - bx^2)^m}$.

37. $\log \sqrt{\frac{y^2 + 1}{y^2 - 1}}$.

38. $e^{\arcsec 2\theta}$.

39. $\sqrt{\frac{(2 - 3x)^3}{1 + 4x}}$

40. $\frac{\sqrt[3]{a^2 - x^2}}{\cos x}$.

41. $e^x \log \sin x$.

42. $/arcsin \frac{x}{\sqrt{1 + x^2}}$.

43. $\arctan a^x$.

44. $a^{\sin^2 mx}$.

45. $\cot^3(\log ax)$.

46. $(1 - 3x^2)e^{\frac{1}{x}}$.

47. $\log \frac{\sqrt{1 - x^2}}{\sqrt[3]{1 + x^3}}$.

1. As was pointed out on p. 44 [§ 42], it might be possible to eliminate $v$ between the two given expressions so as to find $y$ directly as a function of $x$, but in most cases the above method is to be preferred.
2. This process will be justified in §127. Only corresponding values of $x$ and $y$ which satisfy the given equation may be substituted in the derivative.