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(For convex curveswe shall give a separate discussion.)Let the function be given by!x 2 sin x1 , if x = 0,f (x) =0if x = 0.The graph of this function is shown by the thick line in Fig. 5.5.Let us find the tangent to the graph at the point (0, 0). Sincex 2 sin x1 − 01= lim x sin = 0,x→0x→0x −0xf (0) = limthe tangent has the equation y − 0 = 0 · (x − 0), or simply y = 0.Thus, in this example the tangent is the x-axis, which the graph intersects infinitely many times in any neighborhood of the point of tangency.By the definition of differentiability of a function f : E → R at a point x0 ∈ E,we havef (x) − f (x0 ) = A(x0 )(x − x0 ) + o(x − x0 )as x → x0 , x ∈ E.Since the right-hand side of this equality tends to zero as x → x0 , x ∈ E, itfollows that limEx→x0 f (x) = f (x0 ), so that a function that is differentiable at apoint is necessarily continuous at that point.We shall show that the converse, of course, is not always true.5.1 Differentiable Functions187Fig.
5.5Fig. 5.6Example 8 Let f (x) = |x|, (Fig. 5.6). Then at the point x0 = 0 we havelimx→x0 −0f (x) − f (x0 )|x| − 0−x= lim= −1,= limx→−0 x − 0x→−0 xx − x0f (x) − f (x0 )|x| − 0x= lim= 1.= limx→x0 +0x→+0 x − 0x→+0 xx − x0limConsequently, at this point the function has no derivative and hence is not differentiable at the point.Example 9 We shall show that ex+h − ex = ex h + o(h) as h → 0.Thus, the function exp(x) = ex is differentiable and d exp(x)h = exp(x)h, orxxxde = ex dx, and therefore exp x = exp x, or dedx = e .Proofex+h − ex = ex eh − 1 = ex h + o(h) = ex h + o(h).Here we have used the formula eh − 1 = h + o(h) obtained in Example 39 ofSect.
3.2.4.Example 10 If a > 0, then a x+h − a x = a h (ln a)h + o(h) as h → 0. Thus da x =xxa x (ln a) dx and dadx = a ln a.1885Differential CalculusProofa x+h − a x = a x a h − 1 = a x eh ln a − 1 == a x h ln a + o(h ln a) = a x (ln a)h + o(h)as h → 0.Example 11 If x = 0, then ln |x + h| − ln |x| = x1 h + o(h) as h → 0.
Thus d ln |x| =d ln |x|11x dx and dx = x .Proofln |x + h| − ln |x| = ln1 +h .xFor |h| < |x| we have |1 + hx | = 1 + hx , and so for sufficiently small values of h wecan write hh1h= +o= h + o(h)ln |x + h| − ln |x| = ln 1 +xxxxas h → 0. Here we have used the relation ln(1 + t) = t + o(t) as t → 0, shown inExample 38 of Sect. 3.2.4.Example 12 If x = 0 and 0 < a = 1, then loga |x + h| − loga |x| =h → 0. Thus, d loga |x| =lx ln adx andd loga |x|dx=lx ln a h + o(h)aslx ln a .Proofh h=log1+=axx h1 hhl1ln 1 +=+o=h + o(h).=ln axln a xxx ln aloga |x + h| − loga |x| = loga 1 +Here we have used the formula for transition from one base of logarithms toanother and the considerations explained in Example 11.5.1.6 Problems and Exercises1.
Show thata) the tangent to the ellipsex2 y2+=1a 2 b25.1 Differentiable Functions189at the point (x0 , y0 ) has the equationxx0 yy0+ 2 = 1;a2b√22√b) light rays from a source located at a focus F1 = (− a − b , 0) or F2 =22( a − b , 0) of an ellipse with semiaxes a > b > 0 are gathered at the other focusby an elliptical mirror.2. Write the formulas for approximate computation of the following values:a)b)c)d)sin( π6 + α) for values of α near 0;sin(30◦ + α ◦ ) for values of α ◦ near 0;cos( π4 + α) for values of α near 0;cos(45◦ + α ◦ ) for values of α ◦ near 0.3. A glass of water is rotating about its axis at constant angular velocity ω.
Lety = f (x) denote the equation of the curve obtained by cutting the surface of theliquid with a plane passing through its axis of rotation.2a) Show that f (x) = ωg x, where g is the acceleration of free fall. (See Example 5.)b) Choose a function f (x) that satisfies the condition given in part a). (See Example 6.)c) Does the condition on the function f (x) given in part a) change if its axis ofrotation does not coincide with the axis of the glass?4. A body that can be regarded as a point mass is sliding down a smooth hill underthe influence of gravity. The hill is the graph of a differentiable function y = f (x).a) Find the horizontal and vertical components of the acceleration vector thatthe body has at the point (x0 , y0 ).b) For the case f (x) = x 2 when the body slides from a great height, find thepoint of the parabola y = x 2 at which the horizontal component of the accelerationis maximal.5.
Set!Ψ0 (x) =x,if 0 ≤ x ≤ 12 ,1 − x, if12≤ x ≤ 1,and extend this function to the entire real line so as to have period 1. We denote theextended function by ϕ0 . Further, letϕn (x) =1 n ϕ0 4 x .4n1905Differential CalculusThe function ϕn has period 4−n and a derivative equal to +1 or −1 everywherekexcept at the points x = 22n+1, k ∈ Z. Letf (x) =∞ϕn (x).n=1Show that the function f is defined and continuous on R, but does not have aderivative at any point. (This example is due to the well-known Dutch mathematician B.L.
van der Waerden (1903–1996). The first examples of continuous functionshaving no derivatives were constructed by Bolzano (1830) and Weierstrass (1860).)5.2 The Basic Rules of DifferentiationConstructing the differential of a given function or, equivalently, the process of finding its derivative, is called differentiation.75.2.1 Differentiation and the Arithmetic OperationsTheorem 1 If functions f : X → R and g : X → R are differentiable at a pointx ∈ X, thena) their sum is differentiable at x, and(f + g) (x) = f + g (x);b) their product is differentiable at x, and(f · g) (x) = f (x) · g(x) + f (x) · g (x);c) their quotient is differentiable at x if g(x) = 0, and f (x)g(x) − f (x)g (x)f.(x) =gg 2 (x)Proof In the proof we shall rely on the definition of a differentiable function andthe properties of the symbol o(·) proved in Sect.
3.2.4.7 Although the problems of finding the differential and finding the derivative are mathematicallyequivalent, the derivative and the differential are nevertheless not the same thing. For that reason,for example, there are two terms in French – dérivation, for finding the derivative, and différentiation, for finding the differential.5.2 The Basic Rules of Differentiation191a) (f + g)(x + h) − (f + g)(x) = = f (x + h) + g(x + h) − f (x) + g(x) = = f (x + h) − f (x) + g(x + h) − g(x) = = f (x)h + o(h) + g (x)h + o(h) = f (x) + g (x) h + o(h) == f + g (x)h + o(h).b) (f · g)(x + h) − (f · g)(x) == f (x + h)g(x + h) − f (x)g(x) == f (x) + f (x)h + o(h) g(x) + g (x)h + o(h) − f (x)g(x) == f (x)g(x) + f (x)g (x) h + o(h).c) Since a function that is differentiable at a point x ∈ X is continuous at thatpoint, taking account of the relation g(x) = 0 and the properties of continuous functions, we can guarantee that g(x + h) = 0 for sufficiently small values of h.
In thefollowing computations it is assumed that h is small: ff(x + h) −(x) =ggf (x + h) f (x)1−=f (x + h)g(x) − f (x)g(x + h) =g(x + h)g(x)g(x)g(x + h)1=+o(1)f (x) + f (x)h + o(h) g(x) −2g (x)− f (x) g(x) + g (x)h + o(h) =1=+ o(1) f (x)g(x) − f (x)g (x) h + o(h) =g 2 (x)==f (x)g(x) − f (x)g (x)h + o(h).g 2 (x)Here we have used the continuity of g at the point x and the relation g(x) = 0 todeduce that11= 2 ,limh→0 g(x)g(x + h)g (x)that is,11= 2+ o(1),g(x)g(x + h) g (x)where o(1) is infinitesimal as h → 0, x + h ∈ X.Corollary 1 The derivative of a linear combination of differentiable functionsequals the same linear combination of the derivatives of these functions.1925Differential CalculusProof Since a constant function is obviously differentiable and has a derivativeequal to 0 at every point, taking f ≡ const = c in statement b) of Theorem 1, wefind (cg) (x) = cg (x).Now, using statement a) of Theorem 1, we can write(c1 f + c2 g) (x) = (c1 f ) (x) + (c2 g) (x) = c1 f (x) + c2 g (x).Taking account of what has just been proved, we verify by induction that(c1 f1 + · · · + cn fn ) (x) = c1 f1 (x) + · · · + cn fn (x).Corollary 2 If the functions f1 , .
. . , fn are differentiable at x, then(f1 · · · fn ) (x) = f1 (x)f2 (x) · · · fn (x) ++ f1 (x)f2 (x)f3 (x) · · · fn (x) + · · · + f1 (x) · · · fn−1 (x)fn (x).Proof For n = 1 the statement is obvious.If it holds for some n ∈ N, then by statement b) of Theorem 1 it also holds for(n + 1) ∈ N.
By the principle of induction, we conclude that the formula is valid forany n ∈ N.Corollary 3 It follows from the relation between the derivative and the differentialthat Theorem 1 can also be written in terms of differentials. To be specific:a) d(f + g)(x) = df (x) + dg(x);b) d(f · g)(x) = g(x) df (x) + f (x) dg(x);(x) dg(x)if g(x) = 0.c) d( fg )(x) = g(x) df (x)−fg 2 (x)Proof Let us verify, for example, statement a).d(f + g)(x)h = (f + g) (x)h = f + g (x)h == f (x) + g (x) h = f (x)h + g (x)h == df (x)h + dg(x)h = df (x) + dg(x) h,and we have verified that d(f + g)(x) and df (x) + dg(x) are the same function. Example 1 (Invariance of the definition of velocity) We are now in a position toverify that the instantaneous velocity vector of a point mass defined in Sect.
5.1.1is independent of the Cartesian coordinate system used to define it. In fact we shallverify this for all affine coordinate systems.Let (x 1 , x 2 ) and (x̃ 1 , x̃ 2 ) be the coordinates of the same point of the plane in twodifferent coordinate systems connected by the relationsx̃ 1 = a11 x 1 + a21 x 2 + b1 ,x̃ 2 = a12 x 1 + a22 x 2 + b2 .(5.26)5.2 The Basic Rules of Differentiation193Since any vector (in affine space) is determined by a pair of points and its coordinates are the differences of the coordinates of the terminal and initial points ofthe vector, it follows that the coordinates of a given vector in these two coordinatesystems must be connected by the relationsṽ 1 = a11 v 1 + a21 v 2 ,(5.27)ṽ 2 = a12 v 1 + a22 v 2 .If the law of motion of the point is given by functions x 1 (t) and x 2 (t) in onesystem of coordinates, it is given in the other system by functions x̃ 1 (t) and x̃ 2 (t)connected with the first set by relations (5.26).Differentiating relations (5.26) with respect to t, we find by the rules for differentiation1x̃˙ = a11 ẋ 1 + a21 ẋ 2 ,(5.28)2x̃˙ = a12 ẋ 1 + a22 ẋ 2 .Thus the coordinates (v 1 , v 2 ) = (ẋ 1 , ẋ 2 ) of the velocity vector in the first system1 2and the coordinates (ṽ 1 , ṽ 2 ) = (x̃˙ , x̃˙ ) of the velocity vector in the second systemare connected by relations (5.27), telling us that we are dealing with two differentexpressions for the same vector.Example 2 Let f (x) = tan x.
We shall show that f (x) = cos12 x at every point wheresin xcos x = 0, that is, in the domain of definition of the function tan x = cosx.It was shown in Examples 1 and 2 of Sect. 5.1 that sin (x) = cos x and cos x =− sin x, so that by statement c) of Theorem 1 we find, when cos x = 0,tan x ==sincos(x) =sin x cos x − sin x cos x=cos2 xcos x cos x + sin x sin x1=.2cos xcos2 xExample 3 cot x = − 12 wherever sin x = 0, that is, in the domain of definitionsin xxof cot x = cossin x .Indeed,cos cos x sin x − cos x sin xcot x =(x) ==sinsin2 x=− sin x sin x − cos x cos xsin2 x=−1sin2 x.Example 4 If P (x) = c0 + c1 x + · · · + cn x n is a polynomial, then P (x) = c1 +2c2 x + · · · + ncn x n−1 .1945Indeed, since dxdx = 1, by Corollary 2 we havenow follows from Corollary 1.dx ndxDifferential Calculus= nx n−1 , and the statement5.2.2 Differentiation of a Composite Function (Chain Rule)Theorem 2 (Differentiation of a composite function) If the function f : X → Y ⊂R is differentiable at a point x ∈ X and the function g : Y → R is differentiable atthe point y = f (x) ∈ Y , then the composite function g ◦ f : X → R is differentiableat x, and the differential d(g ◦ f )(x) : T R(x) → T R(g(f (x))) of their compositionequals the composition df (y) ◦ df (x) of their differentialsdf (x) : T R(x) → T R y = f (x) and dg y = f (x) : T R(y) → T R g(y) .Proof The conditions for differentiability of the functions f and g have the formf (x + h) − f (x) = f (x)h + o(h) as h → 0, x + h ∈ X,g(y + t) − g(y) = g (y)t + o(t) as t → 0, y + t ∈ Y.We remark that in the second equality here the function o(t) can be consideredto be defined for t = 0, and in the representation o(t) = γ (t)t, where γ (t) → 0 ast → 0, y + t ∈ Y , we may assume γ (0) = 0.
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