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

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EXAMPLES

Differentiate the following:

1. y=arctanax2.
Solution. dydx =ddx(ax2)1+(ax2)2 by XX
[v=ax2.]
  =2ax1+a2x4.
2. y=arcsin(3x4x3).
Solution. dydx ddx(3x4x31(3x4x3)2 by XVIII
[v=3x4x3.]
  312x219x2+24x416x6=31x2.
3. arcsecx2+1x21
Solution. dydx =ddx(x2+1x21)x2+1x21(x2+1x21)21 by XXII
v=[x2+1x21].
  =(x21)2x(x2+1)2x(x21)2x2+1x212xx21=2x2+1

4. ddxarcsinxa=1a2x2.

5. ddxarccot(x25)=2x1+(x25)2.

6. ddxarctan2x1x2=21+x2.

7. ddxarccsc12x21=21x2.

8. ddxarcvers2x2=21x2.

9. ddxarctan1x=121x(2x).

10. ddxarccsc32x=294x2.

11. ddxarcvers2x21+x2=21+x2.

12. ddxarctanxa=aa2+x2.

13. ddxarcsinx+12=112xx2.

14. f(x)=xa2x$2+a2arcsinxa  f(x) = 2a2x2.
15. f(x)=a2x2+aarcsinxa  f(x) =(axa+x)12
16. x=rarcversyr2ryy2. dxdy =y2ryy2.
17. θ=arcsin(3r1)  dθdr =36r9r2.
18. ϕ=arctanr+a1ar dϕdr =11+r2.
19. s=arcsec11t2. dsdt 11t2.

20. ddx(xarcsinx)=arcsinx+x1x2.

21. ddθ(tanθarctanθ)=sec2θarctanθtanθ1+θ2.

22. ddt[log(arccost)]=1arccost1t2.

23. f(y)=arccos(logy) .  f(y) =1y1(logy)2.
24. f(θ)=arcsinsinθ  f(θ) =121+cscθ.
25. f(ϕ)=arctan1cosϕ1+cosϕ.  f(ϕ) =12
26. p=earctanq . dpdq =earctanq1+q2.
27. u=arctanevev2 dudv =2ev+ev
28. s=arccosetetet+et. dsdt =2ev+ev.
29. y=xarcsinx .  y =xarcsinx(arcsinxx+logx1x2).
30. y=exxarctanx  y =exx[11+x2+xxarctanx(1+logx)]
31. y=arcsin(sinx) .  y = 1.
32. y=arctan4sinx3+5cosx.  y =45+3cosx.
33. y=arccotax+logxax+a.  y =2ax2x4a4.
34. y=log(1+x1x)1412arctanx.  y =x21x4.
35. y=1x2arcsinxx.  y =xarcsinx1x2.

36. Differentiate the following functions:

(a) ddxarcsin2x2 (f) ddtt3arcsint3. (k) ddyarcsin1y2.
(b) ddxarctana2x. (g) ddtearctanat. (l) ddzarctan(log3az).
(c) ddxarcsecxa. (h) ddϕtanϕ2arctanϕ12. (m) dds(a2+s2)arcsecs2.
(d) ddxxarccosx. (i) ddθarcsinaθ. (n) ddαarccot2α3.
(e) ddxx2arccotax. (j) ddθarctan1+θ2. (o) ddt1t2arcsint.

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 dydv and dvdx by differentiation and then substitute the results in

dydx =dydvdvdx by XXV

to find dydx.[1]

In general our results should be expressed explicitly in terms of the independent variable; that is, dydx in terms of x,dxdy in terms of y,dϕdθ in terms of θ, etc.

37. y=2v24,v=3x2+1.

dydv=4v;dvdx=6x; substituting in XXV,

dydx=4v6x=24x(3x2+1).

38. y=tan2v,v=arctan(2x1).

dydv=2sec22v;dvdx=12x22x+1; substituting in XXV,

dydx=2sec22v2x22x+1=2tan22v+12x22x+1=2x22x+12(xx2)2

[Since v=arctan(2x1),tanv=2x1,tan2v=2x12x2x2.]

39. y=3v24v+5,v=2x35 . dydx = 72x5204x2.
40. y=2v3v2,v=x2x1. dydx =4(x2)2
41. y=log(a2v2) . dydx = 2tanx.
42. y=arctan(a+v),v=ex  dydx = ex1+(a+ex)2.
43. r=e2s+es,s=log(tt2) . drdt = 4t36t2+1.

In the following examples first find dxdy by differentiation and then substitute in

dydx = 1dxdy by XXVI

to find dydx.

44. x=y1+y. dydx = 21+y2+3y=2x2y+3y2.
45. x=1+cosy. dydx = 21+cosysiny=22x2.
46. x=y1+logy. dydx = (1+logy)2logy.
47. x=aloga+a2y2y. dydx = ya2y2a2.
48. x=rarcversyr2ryy2. dydx = 2ryy.

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

Template:Anchor+ 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

x24y=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,

x2+y2+z2a2=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 =x24
and  y =±a2x2z2;

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) ax6+2x3yy7x10=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) ddxf(x,y)=0.

Let us apply this rule in finding dydx from (B).

ddx(ax6+2x3yy7x10)=0; by (C)
ddx(ax6)+ddx(2x3y)ddx(y7x)ddx(10)=0;
6ax5+2x3dydx+6x2yy77xy6dydx=0;
(2x37xy6)dydx=y76ax56x2y;
dydx=y76ax56x2y2x37xy6 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. y2=4px . dydx = 2py.
2. x2+y2=r2 . dydx = xy.
3. b2x2+a2y2=a2b2 . dydx = b2xa2y.
4. y33y+2ax=0 . dydx = 2a3(1y2).
5. x12+y12=a12 . dydx = yx.
6. x23+y23=a23. dydx = yx3.
7. (xa)2+(yb)23=1 . dydx = 3b23xy13a2.
8. y22xy+b2=0 . dydx = yyx
9. x3+y33axy=0 . dydx = ayx2y2ax.
10. xy=yx . dydx = y2xylogyx2xylogx.
11. ρ2=a2cos2θ . dρdθ = a2sin2θρ.
12. ρ2cosθ=a2sin3θ . dρdθ = 3a2cos3θ+ρ2sinθ2ρcosθ.
13. cos(uv)=cv. dudv = c+usin(uv)vsin(uv)
14. θ=cos(θ+ϕ) . dθdϕ = sin(θ+ϕ)1+sin(θ+ϕ).

15. Find dydx from the following equations:

(a) x2=ay. (f) xy+y2+4x=0. (k) tanx+y3=0.
(b) x2+4y2=16. (g) yx2y3=5. (l) cosy+3x2=0.
(c) b2x2a2y2=a2b2. (h) x22x3y3. (m) xcoty+y=0.
(d) y2=x3+a. (i) x2y3+4y=0. (n) y2=logx.
(e) x2y2=16. (j) y2=sin2x. (o) ex2+2y3=0.

16. A race track has the form of the circle x2+y2=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 252 rods east of OY? Ans. Southeast or southwest.
(b) when 252 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. arcsin14x2 . 214x2 Ans.
2. xex2. ex2(2x2+1).
3. logsinv2. 12cotv2.
4. arccosay. ayy2a2.
5. xa2x2. a2(a2x2)32.
6. x1+logx. logx(1+logx)2
7. logsec(12x). 2tan(12x).
8. x2e23x. xe23x(23x).
9. log1cost1+cost csct
10. arcsin12(1cosx). 12
11. arctan2ss21. 2(15s2)s21.
12. (2x1)21+x3. 7+4x3(1+x)21+x3.
13. x3arcsinx3+(x2+2)1x29. x2arcsinx.

14. tan2θ3+logsec2θ3.

15. arctan12(e2x+e2x).

16. (3x)2x.

17. xtanx.

18. (x+2)13(x21)25x32.

19. esec(13x).

20. arctan1x2.

21. z2cosz.

22. etanx2.

23. logsin212θ.

24. eaxlogsinax.

25. sin3ϕcosϕ.

26. a2(bcxn)m.

27. m+x1+m2emarctanx1+x2.

28. tan2xlogsec2x.

29. 3log(2cosx+3sinx)+2x13.

30. arccotax+logxax+a.

31. (logtan3x2)3.

32. 23t12+4t13+t2t.

33. (1+x)(12x)(2+x)(3+x)(23x).

34. arctan(log3x).

35. (baxm)n3.

36. log(a2bx2)m.

37. logy2+1y21.

38. earcsec2θ.

39. (23x)31+4x

40. a2x23cosx.

41. exlogsinx.

42. /arcsinx1+x2.

43. arctanax.

44. asin2mx.

45. cot3(logax).

46. (13x2)e1x.

47. log1x21+x33.

Template:Rule

  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.