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Setting f (x) = y and f (x + h) = y + t,by the differentiability (and hence continuity) of f at the point x we conclude thatt → 0 as h → 0, and if x + h ∈ X, then y + t ∈ Y . By the theorem on the limit of acomposite function, we now haveγ f (x + h) − f (x) = α(h) → 0 as h → 0, x + h ∈ X,and thus if t = f (x + h) − f (x) theno(t) = γ f (x + h) − f (x) f (x + h) − f (x) == α(h) f (x)h + o(h) = α(h)f (x)h + α(h)o(h) == o(h) + o(h) = o(h)as h → 0, x + h ∈ X,(g ◦ f )(x + h) − (g ◦ f )(x) = g f (x + h) − g f (x) == g(y + t) − g(y) = g (y)t + o(t) == g f (x) f (x + h) − f (x) + o f (x + h) − f (x) == g f (x) f (x)h + o(h) + o f (x + h) − f (x) == g f (x) f (x)h + g f (x) o(h) + o f (x + h) − f (x) .Since we can interpret the quantity g (f (x))(f (x)h) as the value dg(f (x)) ◦df (x)h of the composition hdg(y)◦df (x) −→ g (f (x))df (x)· f (x)h of the mappings h →5.2 The Basic Rules of Differentiation195dg(y)f (x)h, τ → g (y)τ at the displacement h, to complete the proof it remains onlyfor us to remark that the sumg f (x) o(h) + o f (x + h) − f (x)is infinitesimal compared with h as h → 0, x + h ∈ X, or, as we have already established,o f (x + h) − f (x) = o(h) as h → 0, x + h ∈ X.Thus we have proved that(g ◦ f )(x + h) − (g ◦ f )(x) == g f (x) · f (x)h + o(h)as h → 0, x + h ∈ X.Corollary 4 The derivative (g ◦ f ) (x) of the composition of differentiable realvalued functions equals the product g (f (x)) · f (x) of the derivatives of these functions computed at the corresponding points.There is a strong temptation to give a short proof of this last statement in Leibniz’notation for the derivative, in which if z = z(y) and y = y(x), we havedzdz dy=· ,dx dy dxdzwhich appears to be completely natural, if one regards the symbol dyor dydx not as aunit, but as the ratio of dz to dy or dy to dx.The idea for a proof that thereby arises is to consider the difference quotientΔzΔz Δy=·Δx Δy Δxand then pass to the limit as Δx → 0.
The difficulty that arises here (which we alsohave had to deal with in part!) is that Δy may be 0 even if Δx = 0.Corollary 5 If the composition (fn ◦ · · · ◦ f1 )(x) of differentiable functions y1 =f1 (x), . . . , yn = fn (yn−1 ) exists, then(fn ◦ · · · ◦ f1 ) (x) = fn (yn−1 )fn−1(yn−2 ) · · · f1 (x).Proof The statement is obvious if n = 1.If it holds for some n ∈ N, then by Theorem 2 it also holds for n + 1, so that bythe principle of induction, it holds for any n ∈ N.Example 5 Let us show that for α ∈ R we havethat is, dx α = αx α−1 dx anddx αdx(x + h)α − x α = αx α−1 h + o(h)= αx α−1 in the domain x > 0,as h → 0.1965Differential CalculusProof We write x α = eα ln x and apply the theorem, taking account of the results ofExamples 9 and 11 from Sect.
5.1 and statement b) of Theorem 1.Let g(y) = ey and y = f (x) = α ln(x). Then x α = (g ◦ f )(x) and(g ◦ f ) (x) = g (y) · f (x) = ey ·αα= eα ln x · = αx α−1 .xxExample 6 The derivative of the logarithm of the absolute value of a differentiablefunction is often called its logarithmic derivative.Since F (x) = ln |f (x)| = (ln ◦| | ◦ f )(x), by Example 11 of Sect. 5.1, we have (x).F (x) = (ln |f |) (x) = ff (x)Thusf (x)df (x)d ln |f | (x) =dx =.f (x)f (x)Example 7 (The absolute and relative errors in the value of a differentiable functioncaused by errors in the data for the argument) If the function f is differentiableat x, thenf (x + h) − f (x) = f (x)h + α(x; h),where α(x; h) = o(h) as h → 0.Thus, if in computing the value f (x) of a function, the argument x is determinedwith absolute error h, the absolute error |f (x + h) − f (x)| in the value of the function due to this error in the argument can be replaced for small values of h by theabsolute value of the differential |df (x)h| = |f (x)h| at displacement h.
(x)h|(x)h|= |dfThe relative error can then be computed as the ratio |ff (x)||f (x)| or as the(x)||h| of the logarithmic derivative of the functionabsolute value of the product | ff (x)and the magnitude of the absolute error in the argument.We remark by the way that if f (x) = ln x, then d ln x = dxx , and the absolute errorin determining the value of a logarithm equals the relative error in the argument.This circumstance can be beautifully exploited for example, in the slide rule (andmany other devices with nonuniform scales).
To be specific, let us imagine that witheach point of the real line lying right of zero we connect its coordinate y and writeit down above the point, while below the point we write the number x = ey . Theny = ln x. The same real half-line has now been endowed with a uniform scale y anda nonuniform scale x (called logarithmic). To find ln x, one need only set the cursoron the number x and read the corresponding number y written above it. Since theprecision in setting the cursor on a particular point is independent of the number xor y corresponding to it and is measured by some quantity Δy (the length of theinterval of possible deviation) on the uniform scale, we shall have approximatelythe same absolute error in determining both a number x and its logarithm y; andin determining a number from its logarithm we shall have approximately the samerelative error in all parts of the scale.5.2 The Basic Rules of Differentiation197Example 8 Let us differentiate a function u(x)v(x) , where u(x) and v(x) are differentiable functions and u(x) > 0.
We write u(x)v(x) = ev(x) ln u(x) and use Corollary 5. Thenu (x)dev(x) ln u(x)= ev(x) ln u(x) v (x) ln u(x) + v(x)=dxu(x)= u(x)v(x) · v (x) ln u(x) + v(x)u(x)v(x)−1 · u (x).5.2.3 Differentiation of an Inverse FunctionTheorem 3 (The derivative of an inverse function) Let the functions f : X → Y andf −1 : Y → X be mutually inverse and continuous at points x0 ∈ X and f (x0 ) =y0 ∈ Y respectively. If f is differentiable at x0 and f (x0 ) = 0, then f −1 is alsodifferentiable at the point y0 , and −1 −1f(y0 ) = f (x0 ) .Proof Since the functions f : X → Y and f −1 : Y → X are mutually inverse, thequantities f (x) − f (x0 ) and f −1 (y) − f −1 (y0 ), where y = f (x), are both nonzeroif x = x0 .
In addition, we conclude from the continuity of f at x0 and f −1 at y0 that(X x → x0 ) ⇔ (Y y → y0 ). Now using the theorem on the limit of a compositefunction and the arithmetic properties of the limit, we findlimY y→y0f −1 (y) − f −1 (y0 )x − x0== limXx→x0 f (x) − f (x0 )y − y0=limXx→x01(x0 )( f (x)−f)x−x0=1.f (x0 )Thus we have shown that the function f −1 : Y → X has a derivative at y0 andthat−1 −1 (y0 ) = f (x0 ) .fRemark 1 If we knew in advance that the function f −1 was differentiable at y0 ,we would find immediately by the identity (f −1 ◦ f )(x) = x and the theorem ondifferentiation of a composite function that (f −1 ) (y0 ) · f (x0 ) = 1.Remark 2 The condition f (x0 ) = 0 is obviously equivalent to the statement thatthe mapping h → f (x0 )h realized by the differential df (x0 ) : T R(x0 ) → T R(y0 )has the inverse mapping [df (x0 )]−1 : T R(y0 ) → T R(x0 ) given by the formula τ →(f (x0 ))−1 τ .Hence, in terms of differentials we can write the second statement in Theorem 3as follows:1985Differential CalculusIf a function f is differentiable at a point x0 and its differential df (x0 ) :T R(x0 ) → T R(y0 ) is invertible at that point, then the differential of the functionf −1 inverse to f exists at the point y0 = f (x0 ) and is the mapping)*−1df −1 (y0 ) = df (x0 ): T R(y0 ) → T R(x0 ),inverse to df (x0 ) : T R(x0 ) → T R(y0 ).Example 9 We shall show that arcsin y = √ 11−y 2for |y| < 1.
The functions sin :[−π/2, π/2] → [−1, 1] and arcsin : [−1, 1] → [−π/2, π/2] are mutually inverseand continuous (see Example 8 of Sect. 4.2) and sin (x) = cos x = 0 if |x| < π/2.For |x| < π/2 we have |y| < 1 for the values y = sin x.
Therefore, by Theorem 3arcsin y =1111=#.==#2sin x cos x1 − y21 − sin xThe sign in front of the radical is chosen taking account of the inequality cos x > 0for |x| < π/2.Example 10 Reasoning as in the preceding example, one can show (taking accountof Example 9 of Sect. 4.2) thatarccos y = − #11 − y2for |y| < 1.Indeed,arccos y =1111= −#.=−= −√2cos xsin x1 − cos x1 − y2The sign in front of the radical is chosen taking account of the inequality sin x > 0if 0 < x < π .Example 11 arctan y =Indeed,arctan y =1,1+y 2y ∈ R.1111==.= cos2 x =12tan x ( 2 )1 + tan x 1 + y 2cos x1Example 12 arccot y = − 1+y2 , y ∈ R.Indeedarccot y =1111==−.= − sin2 x = −2cot x (− 12 )1 + cot x1 + y2sin x5.2 The Basic Rules of Differentiation199Example 13 We already know (see Examples 10 and 12 of Sect.
5.1) that the functions y = f (x) = a x and x = f −1 (y) = loga y have the derivatives f (x) = a x ln aand (f −1 ) (y) = y lnl a .Let us see how this is consistent with Theorem 3: −1 f(y) =f (x) =1f (x)=ax1(f −1 ) (y)1l=,ln a y ln a=1( y lnl a )= y ln a = a x ln a.Example 14 The hyperbolic and inverse hyperbolic functions and their derivatives.The functions1 xe − e−x ,21cosh x = ex + e−x2sinh x =are called respectively the hyperbolic sine and hyperbolic cosine8 of x.These functions, which for the time being have been introduced purely formally,arise just as naturally in many problems as the circular functions sin x and cos x.We remark thatsinh(−x) = − sinh x,cosh(−x) = cosh x,that is, the hyperbolic sine is an odd function and the hyperbolic cosine is an evenfunction.Moreover, the following basic identity is obvious:cosh2 x − sinh2 x = 1.The graphs of the functions y = sinh x and y = cosh x are shown in Fig.
5.7.It follows from the definition of sinh x and the properties of the function ex thatsinh x is a continuous strictly increasing function mapping R in a one-to-one manneronto itself. The inverse function to sinh x thus exists, is defined on R, is continuous,and is strictly increasing.This inverse is denoted arsinh y (read “area-sine of y”).9 This function is easilyexpressed in terms of known functions.
In solving the equation1 xe − e−x = y28 From9 Thethe Latin phrases sinus hyperbolici and cosinus hyperbolici.full name is area sinus hyperbolici (Lat.); the reason for using the term area here instead ofarc, as in the case of the circular functions, will be explained later.2005Differential CalculusFig. 5.7for x, we find successivelye =y+x(ex > 0, and so ex = y −#1 + y21 + y 2 ) andx = ln y + 1 + y 2 .Thus,arsinh y = ln y + 1 + y 2 ,y ∈ R.Similarly, using the monotonicity of the function y = cosh x on the two intervals R− = {x ∈ R | x ≤ 0} and R+ = {x ∈ R | x ≥ 0}, we can construct functionsarcosh− y and arcosh+ y, defined for y ≥ 1 and inverse to the function cosh x onR− and R+ respectively.They are given by the formulasarcosh− y = ln y − y 2 − 1 ,arcosh+ y = ln y + y 2 − 1 .From the definitions given above, we find1 xe + e−x = cosh x,21cosh x = ex − e−x = sinh x,2sinh x =and by the theorem on the derivative of an inverse function, we findarsinh y =arcosh− y =1111=#=#, =sinh x cosh x1 + y21 + sinh2 x1111= #,= −# =cosh x sinh x − cosh2 x − 1y2 − 1y > 1,5.2 The Basic Rules of Differentiationarcosh+ y =2011111,=#=# =22cosh x sinh xy −1cosh x − 1y > 1.These last three relations can be verified by using the explicit expressions for theinverse hyperbolic functions arsinh y and arcosh y.For example,−1/211#arsinh y =1 + 1 + y2· 2y =2y + 1 + y2#1 + y2 + y11#=· #=#.y + 1 + y21 + y21 + y2Like tan x and cot x one can consider the functionstanh x =sinh xcosh xcosh x,sinh xcoth x =andcalled the hyperbolic tangent and hyperbolic cotangent respectively, and also thefunctions inverse to them, the area tangentartanh y =1 1+yln,2 1−y|y| < 1,arcoth y =1 y +1ln,2 y −1|y| > 1.and the area cotangentWe omit the solutions of the elementary equations that lead to these formulas.By the rules for differentiation we havetanh x ==coth x ==sinh x cosh x − sinh x cosh xcosh2 xcosh x cosh x − sinh x sinh x2cosh x=cosh x sinh x − cosh x sinh xsinh2 xsinh x sinh x − cosh x cosh x2sinh x=1cosh2 x==−1sinh2 xBy the theorem on the derivative of an inverse functionartanh y ==1=tanh x (11 − tanh x211)cosh2 x== cosh2 x =1,1 − y2,|y| < 1,.2025arcoth y =Differential Calculus11= − sinh2 x = =coth x (− 12 )sinh x=−1coth2 x − 1=−1,y2 − 1|y| > 1.The last two formulas can also be verified by direct differentiation of the explicitformulas for the functions artanh y and arcoth y.5.2.4 Table of Derivatives of the Basic Elementary FunctionsWe now write out (see Table 5.1) the derivatives of the basic elementary functionscomputed in Sects.
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