6.2. DIFFERENTIAL CALCULUS FOR FUNCTIONS OF A SINGLE VARIABLE 255 6.2.3-3. Methods for interpreting other indeterminate expressions. 1 ◦ . Expressions of the form 0⋅∞ and ∞–∞ can be reduced to indeterminate expressions 0 0 or ∞ ∞ by means of algebraic transformations, for instance: u(x) ⋅ v(x)= u(x) 1/v(x) transformation rule 0 ⋅ ∞ =⇒ 0 0 , u(x)–v(x)= 1 u(x) – 1 v(x) : 1 u(x)v(x) transformation rule ∞ – ∞ =⇒ 0 0 . 2 ◦ . Indeterminate expressions of the form 1 ∞ , ∞ 0 , 0 0 can be reduced to expressions of the form 0 0 or ∞ ∞ by taking logarithm and using the formulas ln u v = v ln u = ln u 1/v . Example 2. Let us calculate the limit lim x→0 (cos x) 1/x 2 . We have the indeterminate expression 1 ∞ .Wefind that ln lim x→0 (cos x) 1/x 2 = lim x→0 ln(cos x) 1/x 2 = lim x→0 ln cos x x 2 = lim x→0 (ln cos x) (x 2 ) = lim x→0 (– tan x) 2x =– 1 2 . Therefore, lim x→0 (cos x) 1/x 2 = e –1/2 = 1 √ e . 6.2.4. Higher-Order Derivatives and Differentials. Taylor’s Formula 6.2.4-1. Derivatives and differentials of higher orders. The second-order derivative or the second derivative of a function y = f(x) is the derivative of the derivative f (x). The second derivative is denoted by y andalsobyy xx , d 2 y dx 2 , f (x). The derivative of the second derivative of a function y = f(x) is called the third-order derivative, y =(y ) .Thenth-order derivative of the function y = f (x)isdefined as the derivative of its (n – 1)th derivative: y (n) =(y (n–1) ) . The nth-order derivative is also denoted by y (n) x , d n y dx n , f (n) (x). The second-order differential is the differential of the first-order differential, d 2 y = d(dy). If x is the independent variable, then d 2 y = y ⋅ (dx) 2 . In a similar way, one defines differentials of higher orders. 6.2.4-2. Table of higher-order derivatives of some elementary functions. (x a ) (n) = a(a – 1) (a – n + 1)x a–n ,(a x ) (n) =(lna) n a x , (ln x) (n) =(–1) n–1 (n – 1)! 1 x n ,(log a x) (n) =(–1) n–1 (n – 1)! ln a 1 x n , (sin x) (n) =sin x + πn 2 ,(cosx) (n) =cos x + πn 2 , (sinh x) (n) = cosh x if n is odd, sinh x if n is even, (cosh x) (n) = cosh x if n is even, sinh x if n is odd. 256 LIMITS AND DERIVATIVES 6.2.4-3. Rules for calculating higher-order derivatives. 1. Derivative of a sum (difference) of functions: [u(x) v(x)] (n) = u (n) (x) v (n) (x). 2. Derivatives of a function multiplied by a constant: [au(x)] (n) = au (n) (x)(a = const). 3. Derivatives of a product: [u(x)v(x)] = u (x)v(x)+2u (x)v (x)+u(x)v (x), [u(x)v(x)] = u (x)v(x)+3u (x)v (x)+3u (x)v (x)+u(x)v (x), [u(x)v(x)] (n) = n k=0 C k n u (k) (x)v (n–k) (x) (Leibnitz formula), where C k n are binomial coefficients, u (0) (x)=u(x), v (0) (x)=v(x). 4. Derivatives of a composite function: f(u(x)) = f uu (u x ) 2 + f u u xx , f(u(x)) = f uuu (u x ) 3 + 3f uu u x u xx + f u u xxx . 5. Derivatives of a parametrically defined function x = x(t), y = y(t): y = x t y tt – y t x tt (x t ) 3 , y = (x t ) 2 y ttt – 3x t x tt y tt + 3y t (x tt ) 2 – x t y t x ttt (x t ) 5 , y (n) = (y (n–1) ) t x t . 6. Derivatives of an implicit function defined by the equation F (x, y)=0: y = 1 F 3 y –F 2 y F xx + 2F x F y F xy – F 2 x F yy , y = 1 F 5 y –F 4 y F xxx + 3F x F 3 y F xxy – 3F 2 x F 3 y F xyy + F 3 x F y F yyy + 3F 3 y F xx F xy – 3F x F 2 y F xx F yy – 6F x F 2 y F 2 xy – 3F 3 x F 2 yy + 9F 2 x F y F xy F yy , where the subscripts denote the corresponding partial derivatives. 7. Derivatives of the inverse function x = x(y): x yy =– y xx (y x ) 3 , x yyy =– y xxx (y x ) 4 + 3 (y xx ) 2 (y x ) 5 , x (n) y = 1 y x [x (n–1) y ] x . 6.2. DIFFERENTIAL CALCULUS FOR FUNCTIONS OF A SINGLE VARIABLE 257 6.2.4-4. Taylor’s formula. Suppose that in a neighborhood of a point x = a, the function y = f (x)hasderivatives up to the order (n + 1) inclusively. Then for all x in that neighborhood, the following representation holds: f(x)=f (a)+ f (a) 1! (x – a)+ f (a) 2! (x – a) 2 + ···+ f (n) (a) n! (x – a) n + R n (x), (6.2.4.1) where R n (x)istheremainder term in the Taylor formula. The remainder term can be represented in different forms (6.2.4.1): R n (x)=o[(x – a) n ] (Peano), R n (x)= f (n+1) a + k(x – a) (n + 1)! (x – a) n+1 (Lagrange), R n (x)= f (n+1) a + k(x – a) n! (1 – k) n (x – a) n+1 (Cauchy), R n (x)= f (n+1) a + k(x – a) n!p (1 – k) n+1–p (x – a) n+1 (Schl ¨ omilch and Roche), R n (x)= 1 n! x a f (n+1) (t)(x – t) n dt (integral form), where 0 < k < 1 and p > 0; k depends on x, n, and the structure of the remainder term. The remainders in the form of Lagrange and Cauchy can be obtained as special cases of the Schl ¨ omilch formula with p = n + 1 and p = 1, respectively. For a = 0, the Taylor formula (6.2.4.1) turns into f(x)=f(0)+ f (0) 1! x + f (0) 2! x 2 + ···+ f (n) (0) n! x n + R n (x) and is called the Maclaurin formula. The Maclaurin formula for some functions: e x = 1 + x 1! + x 2 2! + x 3 3! + ···+ x n n! + R n (x), sin x = x – x 3 3! + x 5 5! – x 7 7! + ···+(–1) n x 2n+1 (2n + 1)! + R 2n+1 (x), cos x = 1 – x 2 2! + x 4 4! – x 6 6! + ···+(–1) n x 2n (2n)! + R 2n (x). 6.2.5. Extremal Points. Points of Inflection 6.2.5-1. Maximum and minimum. Points of extremum. Let f (x) be a differentiable function on the interval (a, b)andf (x)>0 (resp., f (x)<0) on (a, b). Then f(x)isanincreasing (resp., decreasing) function on that interval*. Suppose that there is a neighborhood of a point x 0 such that for all x ≠ x 0 in that neighborhood we have f(x)>f (x 0 ) (resp., f (x)<f (x 0 )). Then x 0 is called a point of local minimum (resp., local maximum) of the function f(x). Points of local minimum or maximum are called points of extremum. * At some points of the interval, the derivative may vanish. 258 LIMITS AND DERIVATIVES 6.2.5-2. Necessary and sufficient conditions for the existence of extremum. NECESSARY CONDITION OF EXTREMUM. A function f(x) can have an extremum only at points in which its derivative either vanishes or does not exist (or is infinite). FIRST SUFFICIENT CONDITION OF EXTREMUM. Suppose that f(x) is continuous in some neighborhood (x 0 –δ, x 0 +δ) of a point x 0 and differentiable at all points of the neighborhood except, possibly, x 0 .If f (x)>0 for x (x 0 – δ, x 0 ) and f (x)<0 for x (x 0 , x 0 + δ) , then x 0 is a point of local maximum of this function. If f (x)<0 for x (x 0 – δ, x 0 ) and f (x)>0 for x (x 0 , x 0 + δ) ,then x 0 is a point of local minimum of this function. If f (x) is of the same sign for all x ≠ x 0 , x (x 0 – δ, x 0 + δ) ,then x 0 cannot be a point of extremum. SECOND SUFFICIENT CONDITION OF EXTREMUM. Let f(x) be a twice differentiable function in a neighborhood of x 0 . Then the following implications hold: (i) f (x 0 )=0 and f (x 0 )<0 =⇒ f(x) has a local maximum at the point x 0 ; (ii) f (x 0 )=0 and f (x 0 )>0 =⇒ f(x) has a local minimum at the point x 0 . T HIRD SUFFICIENT CONDITION OF EXTREMUM. Let f(x) be a function that is n times differentiable in a neighborhood of a point x 0 and f (x 0 )=f (x 0 )=··· = f (n–1) (x 0 )=0 , but f (n) (x 0 ) ≠ 0 . Then the following implications hold: (i) n is even and f (n) (x 0 )<0 =⇒ f(x) has a local maximum at the point x 0 ; (ii) n is even and f (n) (x 0 )>0 =⇒ f(x) has a local minimum at the point x 0 . If n is odd, then x 0 cannot be a point of extremum. 6.2.5-3. Largest and the smallest values of a function. Let y = f (x) be continuous on the segment [a, b] and differentiable at all points of this segment except, possibly, finitely many points. Then the largest and the smallest values of f(x)on[a, b] belong to the set consisting of f(a), f(b), and the values f(x i ), where x i (a, b) are the points at which f (x) is either equal to zero or does not exist (is infinite). 6.2.5-4. Direction of the convexity of the graph of a function. The graph of a differentiable function y = f(x)issaidtobeconvex upward (resp., convex downward) on the interval (a, b) if for each point of this interval, the graph lies below (resp., above) the tangent line at that point. If the function y = f(x) is twice differentiable on the interval (a, b)andf (x)<0 (resp., f (x)>0), then its graph is convex upward (resp., downward) on that interval. (At some points of the interval, the second derivative may vanish.) Thus, in order to find the intervals on which the graph of a twice differentiable function f(x) is convex upward (resp., downward), one should solve the inequality f (x)<0 (resp., f (x)>0). 6.2.5-5. Inflection points. An inflection point on the graph of a function y = f(x)isdefined as a point (x 0 , f(x 0 )) at which the graph passes from one side of its tangent line to another. At an inflection point, the graph changes the direction of its convexity. 6.2. DIFFERENTIAL CALCULUS FOR FUNCTIONS OF A SINGLE VARIABLE 259 Suppose that the function y = f (x) has a continuous second derivative f (x)insome neighborhood of a point x 0 .Iff (x 0 )=0 and f (x) changes sign as x passes through the point x 0 ,then(x 0 , f(x 0 )) is an inflection point. 6.2.6. Qualitative Analysis of Functions and Construction of Graphs 6.2.6-1. General scheme of analysis of a function and construction of its graph. 1. Determine the domain in which the function is defined. 2. Find the asymptotes of the graph. 3. Find extremal points and intervals of monotonicity. 4. Determine the directions of convexity of the graph and its inflection points. 5. Determine whether the function is odd or even and whether it is periodic. 6. Find the points at which the graph crosses the coordinate axes. 7. Draw the graph, using the properties 1 to 6. Example. Let us examine the function y = ln x x and construct its graph. We use the above general scheme. 1. This function is defined for all x such that 0 < x <+∞. 2. The straight line x = 0 is a vertical asymptote, since lim x→+0 ln x x =–∞.Wefind the oblique asymptotes: k = lim x→+∞ y x = 0, b = lim x→+∞ (y – kx)=0. Therefore, the line y = 0 is a horizontal asymptote of the graph. 3. The derivative y = 1 –lnx x 2 vanishes for ln x = 1. Therefore, the function may have an extremum at x = e.Forx (0, e), we have y > 0, i.e., the function is increasing on this interval. For x (e,+∞), we have y < 0, and therefore the function is decreasing on this interval. At x = e the function attains its maximal value y max = 1 e . One should also examine the points at which the derivative does not exist. There is only one such point, x = 0, and it corresponds to the vertical asymptote (see Item 1). 4. The second derivative y = 2 ln x – 3 x 3 vanishes for x = e 3/2 . On the interval (0, e 3/2 ), we have y < 0, and therefore the graph is convex upward on this interval. For x (e 3/2 ,+∞), we have y > 0, and therefore the graph is convex downward on this interval. The value x = e 3/2 corresponds to an inflection point of the graph, with the ordinate y = 3 2 e –3/2 . 5. This function is neither odd nor even, since it is defined only for x > 0 and the relations f(–x)=f(x) or f(–x)=–f(x) cannot hold. Obviously, this function is nonperiodic. 6. The graph of this function does not cross the y-axis, since for x = 0 the function is undefined. Further, y = 0 only if x = 1, i.e., the graph crosses the x-axis only at the point (1, 0). 7. Using the above results, we construct the graph (Fig. 6.5). 6.2.6-2. Transformations of graphs of functions. Let us describe some methods which in many cases allow us to construct the graph of a function if we have the graph of a simpler function. 1. The graph of the function y = f(x)+a is obtained from that of y = f (x) by shifting the latter along the axis Oy by the distance |a|.Fora > 0 the shift is upward, and for a < 0 downward (see Fig. 6.6 a). 2. The graph of the function y = f(x + a) is obtained from that of y = f (x) by shifting the latter along the Ox by the distance |a|.Fora > 0 the shift is to the left, and for a < 0 to the right (see Fig. 6.6 b). 260 LIMITS AND DERIVATIVES O x y e 3/(2 )e e 10 0.2 1/e 1 20 30 40 3/2 3/2 0.2 0.4 y = ln x x Figure 6.5. Graph of the function y = ln x x . 3. The graph of the function y =–f(x) is obtained from that of y = f(x) by symmetric reflection with respect to the axis Ox (see Fig. 6.6 c). 4. The graph of the function y = f(–x) is obtained from that of y = f (x) by symmetric reflection with respect to the axis Oy (see Fig. 6.6 d). 5. The graph of the function y = kf(x)fork > 1 is obtained from that of y = f(x)by extending the latter k times from the axis Ox,andfor0 < k < 1 by contracting the latter 1/k times to the axis Ox. The points at which the graph crosses the axis Ox remain unchanged (see Fig. 6.6 e). 6. The graph of the function y = f (kx)fork > 1 is obtained from that of y = f(x) by contracting the latter k times to the axis Oy,andfor0 < k < 1 by extending the latter 1/k times from the axis Oy. The points at which the graph crosses the axis Oy remain unchanged (see Fig. 6.6 f). 7. The graph of the function y = |f (x)| is obtained from that of y = f (x) by preserving the parts of the latter for which f (x) ≥ 0 and symmetric reflection, with respect to the axis Ox, of the parts for which f (x)<0 (see Fig. 6.6 g). 8. The graph of the inverse function y = f –1 (x) is obtained from that of y = f(x)by symmetric reflection with respect to the straight line y = x (see Fig. 6.6 h). 6.2.7. Approximate Solution of Equations (Root-Finding Algorithms for Continuous Functions) 6.2.7-1. Preliminaries. For a vast majority of algebraic (transcendental) equations of the form f(x)=0,(6.2.7.1) where f (x) is a continuous function, there are no exact closed-form expressions for the roots. When solving the equation approximately, the first step is to bracket the roots, i.e., find sufficiently small intervals containing exactly one root each. Such an interval [a, b], where the numbers a and b satisfy the condition f(a)f (b)<0 (which is assumed to hold in what follows), can be found, say, graphically. The second step is to compute successive approximations x n [a, b](n = 1, 2, )to the desired root c = lim n→∞ x n , usually by one of the following methods. 6.2. DIFFERENTIAL CALCULUS FOR FUNCTIONS OF A SINGLE VARIABLE 261 O O O O O O O O x x x x x x x x y y y y ()a ()c ()e ()g ()b ()d ( ) f ()h y y y y yfx= |()| yfx= () yfx= () yfx= () yfx= () yfx= () ykfx= () yfkx= () yfx= - () yfx= () yfx= () yfx= () yfxa= (+) yfx a=+() a >0 a >0 k >1 k >1 0< 1k< 0< 1k< a<0 a <0 yfx=-() yfx= () yx= 1 Figure 6.6. Transformations of graphs of functions. 6.2.7-2. Bisection method. To find the root of equation (6.2.7.1) on the interval [a, b], we bisect the interval. If f a + b 2 = 0,thenc = a + b 2 is the desired root. If f a + b 2 ≠ 0, then of the two intervals a, a + b 2 and a + b 2 , b we take the one at whose endpoints the function f(x) has opposite signs. Now we bisect the new, smaller interval, etc. As a result, we obtain either an exact root of equation (6.2.7.1) at some step or an infinite sequence of nested intervals [a 1 , b 1 ], [a 2 , b 2 ], such that f(a n )f(b n )<0. The root is given by the formula . graph of the function y = |f (x)| is obtained from that of y = f (x) by preserving the parts of the latter for which f (x) ≥ 0 and symmetric reflection, with respect to the axis Ox, of the parts for. term. The remainders in the form of Lagrange and Cauchy can be obtained as special cases of the Schl ¨ omilch formula with p = n + 1 and p = 1, respectively. For a = 0, the Taylor formula (6.2.4.1) turns. inflection point. 6.2.6. Qualitative Analysis of Functions and Construction of Graphs 6.2.6-1. General scheme of analysis of a function and construction of its graph. 1. Determine the domain in which