If y f u where u g x , prove that in .NET

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4.13. If y f u where u g x , prove that
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dy dy du assuming that f and g are di erentiable. dx du dx
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Let x be given an increment x 6 0. Then as a consequence u and y take on increments u and y respectively, where y f u u f u ; u g x x g x 1
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Note that as x ! 0, y ! 0 and u ! 0. y dy If u 6 0, let us write  so that  ! 0 as u ! 0 and u du dy y u  u du
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If u 0 for values of x, then (1) shows that y 0 for these values of x. For such cases, we de ne  0. It follows that in both cases, u 6 0 or u 0, (2) holds. Dividing (2) by x 6 0 and taking the limit as x ! 0, we have   dy y dy u u dy u u lim lim  lim  lim lim dx x!0 x x!0 du x x du x!0 x x!0 x!0 x dy du du dy du 0 3 du dx dx du dx
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4.14. Given a
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d d sin x cos x and cos x sin x, derive the formulas dx dx b d 1 sin 1 x p dx 1 x2
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d tan x sec2 x; dx
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  cos x d sin x sin x d cos x d d sin x dx dx tan x dx dx cos x cos2 x cos x cos x sin x sin x 1 2 x cos2 x cos2 x Taking the derivative with respect to x, dy dx or dy 1 1 1 q p dx cos y 2 1 x2 1 sin y
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(b) If y sin 1 x, then x sin y. 1 cos y
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We have supposed here that the principal value =2 @ sin 1 x @ =2, is chosen so that cos y is q q positive, thus accounting for our writing cos y 1 sin2 y rather than cos y 1 sin2 y.
4.15. Derive the formula
d loga e du loga u a > 0; a 6 1 , where u is a di erentiable function of x. dx u dx
dy f u u f u loga u u loga u lim lim u!0 du u!0 u u     1 u u 1 u u= u lim loga lim loga 1 u!0 u u!0 u u u
Consider y f u loga u. By de nition,
DERIVATIVES
[CHAP. 4
Since the logarithm is a continuous function, this can be written ( )   1 u u= u 1 loga lim 1 loga e u!0 u u u by Problem 2.19, 2, with x u= u. d loga e du Then by Problem 4.13, loga u . dx u dx
4.16. Calculate dy=dx if
(a) xy3 3x2 xy 5,
(b) exy y ln x cos 2x.
(a) Di erentiate with respect to x, considering y as a function of x. (We sometimes say that y is an implicit function of x, since we cannot solve explicitly for y in terms of x.) Then d d d d xy3 3x2 xy 5 dx dx dx dx where y 0 dy=dx. b or x 3y2 y 0 y3 1 6x x y 0 y 1 0
Solving, y 0 6x y3 y = 3xy2 x . exy xy 0 y y0 y ln x y 0 2 sin 2x: x
d xy d d e y ln x cos 2x ; dx dx dx Solving;
2x sin 2x xyexy y x2 exy x ln x
4.17. If y cosh x2 3x 1 , nd
(a) dy=dx;
b d 2 y=dx2 .
(a) Let y cosh u, where u x2 3x 1.
Then dy=du sinh u, du=dx 2x 3, and
dy dy du sinh u 2x 3 2x 3 sinh x2 3x 1 dx du dx b      2 d2y d dy d du d2u du sinh u sinh u 2 cosh u 2 dx dx dx dx dx dx dx sinh u 2 cosh u 2x 3 2 2 sinh x2 3x 1 2x 3 2 cosh x2 3x 1
4.18. If x2 y y3 2, nd
(a) y 0 ;
b y 00 at the point 1; 1 .
2xy 1 at 1; 1 2 x2 3xy2
(a) Di erentiating with respect to x, x2 y 0 2xy 3y2 y 0 0 and y0
y 00
  d 0 d 2xy x2 3y2 2xy 0 2y 2xy 2x 6yy 0 y 2 2 dx dx x 3y x2 3y2 2
Substituting x 1, y 1; and y 0 1, we nd y 00 3. 2 8
MEAN VALUE THEOREMS 4.19. Prove Rolle s theorem.
Case 1: f x  0 in a; b . Then f 0 x 0 for all x in a; b . Case 2: f x 6 0 in a; b . Since f x is continuous there are points at which f x attains its maximum and minimum values, denoted by M and m respectively (see Problem 3.34, 3). Since f x 6 0, at least one of the values M; m is not zero. Suppose, for example, M 6 0 and that f  M (see Fig. 4-9). For this case, f  h @ f  .
CHAP. 4]
DERIVATIVES f  h f  @ 0 and h f  h f  @0 lim h!0 h f  h f  A 0 and h f  h f  A0 lim h!0 h
f (x)
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