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5.1 and 5.2.5.2.5 Differentiation of a Very Simple Implicit FunctionLet y = y(t) and x = x(t) be differentiable functions defined in a neighborhoodU (t0 ) of a point t0 ∈ R. Assume that the function x = x(t) has an inverse t = t (x)defined in a neighborhood V (x0 ) of x0 = x(t0 ). Then the quantity y = y(t), whichdepends on t, can also be regarded as an implicit function of x, since y(t) = y(t (x)).Let us find the derivative of this function with respect to x at the point x0 , assumingthat x (t0 ) = 0. Using the theorem on the differentiation of a composite function andthe theorem on differentiation of an inverse function, we obtaindy(t (x)) dy(t) dt (x) yx x=x ==·=0dx x=x0dt t=t0 dx x=x0dy(t)dt |t=t0dx(t)dt |t=t0=yt (t0 ).xt (t0 )(Here we have used the standard notation f (x)|x=x0 := f (x0 ).)If the same quantity is regarded as a function of different arguments, in order toavoid misunderstandings in differentiation, we indicate explicitly the variable withrespect to which the differentiation is carried out, as we have done here.Example 15 (The law of addition of velocities) The motion of a point along a lineis completely determined if we know the coordinate x of the point in our chosencoordinate system (the real line) at each instant t in a system we have chosen formeasuring time.
Thus the pair of numbers (x, t) determines the position of the pointin space and time. The law of motion is written in the form of a function x = x(t).Suppose we wish to express the motion of this point in terms of a different coordinate system (x̃, t˜). For example, the new real line may be moving uniformlywith speed −v relative to the first system. (The velocity vector in this case may beidentified with the single number that defines it.) For simplicity we shall assumethat the coordinates (0,0) refer to the same point in both systems; more precisely,5.2 The Basic Rules of Differentiation203Table 5.1Function f (x)Derivative f (x)1. C (const)02. x ααx α−1x > 0 for α ∈ R3.
a xa x ln ax ∈ R (a > 0, a = 1)4. loga |x|1x ln ax ∈ R\0 (a > 0, a = 1)5. sin xcos x6. cos x− sin x7. tan x1cos2 x− 12sin x√11−x 2−√ 1 21−x11+x21− 1+x2Restrictions on domain of x ∈ Rx ∈ R for α ∈ N8. cot x9. arcsin x10. arccos x11. arctan x12. arccot x13.
sinh xcosh x14. cosh xsinh x15. tanh x1cosh2 x− 12sinh x√11+x 2± √ 12x −1121−x1x 2 −116. coth x√17. arsinh x = ln(x + 1 + x 2 )√18. arcosh x = ln(x ± x 2 − 1)19. artanh x =20. arcoth x =1212lnln1+x1−xx+1x−1x =π2+ πk, k ∈ Zx = πk, k ∈ Z|x| < 1|x| < 1x = 0|x| > 1|x| < 1|x| > 1that at time t˜ = 0 the point x̃ = 0 coincided with the point x = 0 at which the clockshowed t = 0.Then one of the possible connections between the coordinate systems (x, t) and(x̃, t˜) describing the motion of the same point observed from different coordinatesystems is provided by the classical Galilean transformations:x̃ = x + vt,t˜ = t.(5.29)Let us consider a somewhat more general linear connectionx̃ = αx + βt,t˜ = γ x + δt,(5.30)assuming, of course, that this connection is invertible, that is, the determinant of theα βmatrix γ δ is not zero.2045Differential CalculusLet x = x(t) and x̃ = x̃(t˜) be the law of motion for the point under observation,written in these coordinate systems.We remark that, knowing the relation x = x(t), we find by formula (5.30) thatx̃(t) = αx(t) + βt,(5.31)t˜(t) = γ x(t) + δt,and since the transformation (5.30) is invertible, after writingx = α̃ x̃ + β̃ t˜,(5.32)t = γ̃ x̃ + δ̃ t˜,knowing x̃ = x̃(t˜), we findx(t˜) = α̃ x̃(t˜) + β̃ t˜,(5.33)t (t˜) = γ̃ x̃(t˜) + δ̃ t˜.It is clear from relations (5.31) and (5.33) that for the given point there existmutually inverse functions t˜ = t˜(t) and t = t (t˜).We now consider the problem of the connection between the velocitiesV (t) =dx(t)= ẋt (t)dtand Ṽ (t) =dx̃(t˜) ˙= x̃t˜(t˜)dt˜of the point computed in the coordinate systems (x, t) and (x̃, t˜) respectively.Using the rule for differentiating an implicit function and formula (5.31), we havedx̃=dt˜dx̃dtdt˜dt=α dxdt + βγ dxdt + δorṼ (t˜) =αV (t) + β,γ V (t) + δ(5.34)where t˜ and t are the coordinates of the same instant of time in the systems (x, t)and (x̃, t˜).
This is always to be kept in mind in the abbreviated notationṼ =αV + βγV +δ(5.35)for formula (5.34).In the case of the Galilean transformations (5.29) we obtain the classical law ofaddition of velocities from formula (5.35)Ṽ = V + v.(5.36)5.2 The Basic Rules of Differentiation205It has been established experimentally with a high degree of precision (and thisbecame one of the postulates of the special theory of relativity) that in a vacuumlight propagates with a certain velocity c that is independent of the state of motionof the radiating body.
This means that if an explosion occurs at time t = t˜ = 0 atthe point x = x̃ = 0, the light will reach the points x with coordinates such thatx 2 = (ct)2 after time t in the coordinate system (x, t), while in the system (x̃, t˜)this event will correspond to time t˜ and coordinates x̃, where again x̃ 2 = (ct˜)2 .Thus, if x 2 − c2 t 2 = 0, then x̃ 2 − ct˜2 = 0 also, and conversely.
By virtue ofcertain additional physical considerations, one must consider that, in generalx 2 − c2 t 2 = x̃ 2 − c2 t 2 ,(5.37)if (x, t) and (x̃, t˜) correspond to the same event in the different coordinate systemsconnected by relation (5.30). Conditions (5.37) give the following relations on thecoefficients α, β, γ , and δ of the transformation (5.30):α 2 − c2 γ 2 = 1,αβ − c2 γ δ = 0,(5.38)β − c δ = −c .22 22If c = 1, we would have, instead of (5.38),α 2 − γ 2 = 1,γβ= ,δα(5.39)β 2 − δ 2 = −1,from which it follows easily that the general solution of (5.39) (up to a change ofsign in the pairs (α, β) and (γ , δ)) can be given asα = cosh ϕ,γ = sinh ϕ,β = sinh ϕ,δ = cosh ϕ,where ϕ is a parameter.The general solution of the system (5.38) then has the form cosh ϕ c sinh ϕα β= 1γ δcosh ϕc sinh ϕand the transformation (5.30) can be made specific:x̃ = cosh ϕx + c sinh ϕt,t˜ =1sinh ϕx + cosh ϕt.cThis is the Lorentz transformation.(5.40)2065Differential CalculusIn order to clarify the way in which the free parameter ϕ is determined, we recallthat the x̃-axis is moving with speed −v relative to the x-axis, that is, the point x̃ = 0of this axis, when observed in the system (x, t) has velocity −v.
Setting x̃ = 0 in(5.40), we find its law of motion in the system (x, t):x = −c tanh ϕt.Therefore,vtanh ϕ = .(5.41)cComparing the general law (5.35) of transformation of velocities with the Lorentztransformation (5.40), we obtain. = cosh ϕV + c sinh ϕ ,V1c sinh ϕV + cosh ϕor, taking account of (5.41),.= V +v .V1 + vVc2(5.42)Formula (5.42) is the relativistic law of addition of velocities, which for|vV | $ c2 , that is, as c → ∞, becomes the classical law expressed by formula(5.36).The Lorentz transformation (5.40) itself can be rewritten taking account of relation (5.41) in the following more natural form:x + vtx̃ = ,1 − ( vc )2t + v2 x,t˜ = c1 − ( vc )2(5.43)from which one can see that for |v| $ c, that is, as c → ∞, they become the classicalGalilean transformations (5.29).5.2.6 Higher-Order DerivativesIf a function f : E → R is differentiable at every point x ∈ E, then a new functionf : E → R arises, whose value at a point x ∈ E equals the derivative f (x) of thefunction f at that point.The function f : E → R may itself have a derivative (f ) : E → R on E, calledthe second derivative of the original function f and denoted by one of the following5.2 The Basic Rules of Differentiation207two symbols:f (x),d2 f (x),dx 2and if we wish to indicate explicitly the variable of differentiation in the first case, (x).we also write, for example, fxxDefinition By induction, if the derivative f (n−1) (x) of order n − 1 of f has beendefined, then the derivative of order n is defined by the formulaf (n) (x) := f (n−1) (x).The following notations are conventional for the derivative of order n:f (n) (x),dn f (x).dx nAlso by convention, f (0) (x) := f (x).The set of functions f : E → R having continuous derivatives up to ordern inclusive will be denoted C (n) (E, R), and by the simpler symbol C (n) (E), orC n (E, R) and C n (E) respectively wherever no confusion can arise.In particular C (0) (E) = C(E) by our convention that f (0) (x) = f (x).Let us now consider some examples of the computation of higher-order derivatives.Examples16)17)18)19)20)21)f (x)axexsin xcos x(1 + x)αxα22)loga |x|23)ln |x|f (x)a x ln aexcos x− sin xα(1 + x)α−1αx α−1f (x)a x ln2 aex− sin x− cos xα(α − 1)(1 + x)α−2α(α − 1)x α−2·····················f (n) (x)a x lnn aexsin(x + nπ/2)cos(x + nπ/2)α(α − 1) · · · (α − n + 1)(1 + x)α−nα(α − 1) · · · (α − n + 1)x α−n1 −1ln a xx −1−1 −2ln a x(−1)x −2···(−1)n−1 (n−1)! −nxln a(−1)n−1 (n − 1)!x −n···Example 24 (Leibniz’ formula) Let u(x) and v(x) be functions having derivativesup to order n inclusive on a common set E.
The following formula of Leibniz holdsfor the nth derivative of their product:(uv)(n) =n nm=0mu(n−m) v (m) .(5.44)Leibniz’ formula bears a strong resemblance to Newton’s binomial formula, andin fact the two are directly connected.2085Differential CalculusProof For n = 1 formula (5.44) agrees with the rule already established for thederivative of a product.If the functions u and v have derivatives up to order n + 1 inclusive, then, assuming that formula (5.44) holds for order n, after differentiating the left- and right-handsides, we findn n nn(n+1)(n−m+1) (m)uu(n−m) v (m+1) =(uv)=v +mmm=0m=0= u(n+1) v (0) +n k=1=n+1 n+1k=0knn+kk−1u((n+1)−k) v (k) + u(0) v (n+1) =u((n+1)−k) v (k) .Here we have combined the terms containing like productsof the n ofderivativesnfunctions u and v and used the binomial relation k + k−1 = n+1.kThus by induction we have established the validity of Leibniz’ formula.Example 25 If Pn (x) = c0 + c1 x + · · · + cn x n , thenPn (0) = c0 ,Pn (x) = c1 + 2c2 x + · · · + ncn x n−1 and Pn (0) = c1 ,Pn (x) = 2c2 + 3 · 2c3 x + · · · + n(n − 1)cn x n−2 and Pn (0) = 2!c2 ,Pn(3) (x) = 3 · 2c3 + · · · n(n − 1)(n − 2)cn x n−3 and Pn(3) (0) = 3!c3 ,...Pn(n) (x) = n(n − 1)(n − 2) · · · 2cn and Pn(n) (0) = n!cn ,Pn(k) (x) = 0 for k > n.Thus, the polynomial Pn (x) can be written asPn (x) = Pn(0) (0) +1 (1)11Pn (0)x + Pn(2) (0)x 2 + · · · + Pn(n) (0)x n .1!2!n!Example 26 Using Leibniz’ formula and the fact that all the derivatives of a polynomial of order higher than the degree of the polynomial are zero, we can find thenth derivative of f (x) = x 2 sin x: nnsin(n−1) x · 2x +sin(n−2) x · 2 =f (n) (x) = sin(n) (x) · x 2 +12ππ2+ 2nx sin x + (n − 1)+= x sin x + n225.2 The Basic Rules of Differentiation209π+ −n(n − 1) sin x + n=2 2ππ− 2nx cos x + n.= x − n(n − 1) sin x + n22Example 27 Let f (x) = arctan x.
Let us find the values f (n) (0) (n = 1, 2, . . .).1, it follows that (1 + x 2 )f (x) = 1.Since f (x) =1 + x2Applying Leibniz’ formula to this last equality, we find the recursion relation1 + x 2 f (n+1) (x) + 2nxf (n) (x) + n(n − 1)f (n−1) (x) = 0,from which one can successively find all the derivatives of f (x).Setting x = 0, we obtainf (n+1) (0) = −n(n − 1)f (n−1) (0).For n = 1 we find f (2) (0) = 0, and therefore f (2n) (0) = 0. For derivatives of oddorder we havef (2m+1) (0) = −2m(2m − 1)f (2m−1) (0)and since f (0) = 1, we obtainf (2m+1) (0) = (−1)m (2m)!.Example 28 (Acceleration) If x = x(t) denotes the time dependence of a point massmoving along the real line, then dx(t)dt = ẋ(t) is the velocity of the point, and thendẋ(t)dt2= d dtx(t)2 = ẍ(t) is its acceleration at time t.If x(t) = αt + β, then ẋ(t) = α and ẍ(t) ≡ 0, that is, the acceleration in a uniformmotion is zero. We shall soon verify that if the second derivative equals zero, then thefunction itself has the form αt + β.
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