1 Calculus. 1.1 Gradients and the Derivative. Q f(x+h) f(x)

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1 Calculus. Gradients and te Derivative Q f(x+) δy P T δx R f(x) 0 x x+ Let P (x, f(x)) and Q(x+, f(x+)) denote two points on te curve of te function y = f(x) and let R denote te point of intersection of te orizontal troug P wit te vertical troug Q. Let δx = (x + ) x = and δy = f(x + ) f(x). Ten te gradient of te cord P Q is given by QR P R = δy δx f(x + ) f(x) =. Let P T denote te tangent to te curve at P. Te gradient to te curve y = f(x) at P may be approximated by te cord P Q by taking te point Q to be sufficiently close to P. Let Q take a succession of values taking it closer and closer to P ten te straigt line troug P and Q approaces closer and closer to te tangent P T. As Q becomes arbitrarily close to, but never coinciding wit P, te line troug P Q becomes arbitrarily close to te tangent P T. We sall write Q P, and say tat Q tends to P. We sall also say tat Q tends to te limit P. Finally, we sall say tat in te limit as Q tends to P te gradient of te cord P Q tends to te gradient of te tangent P T. Te notion of limit may ten be extended from points to numbers by saying tat in te limit as Q P we ave tat δx 0, 0 and f(x + ) f(x). It is clear now wy we stipulate tat P must not coincide wit Q, for in tis case, δx = 0, = 0, f(x + ) = f(x) and te ratio QR is indeterminate, being of te form 0/0, sometimes called P R an indeterminate form. We now abbreviate te prase in te limit as tends to zero by lim 0 and similarly for δx tending to zero, we write lim δx 0. We define te gradient of te tangent of te curve at te point P to be te limiting value of te gradient of te cord P Q as Q P and we call tis te derivative of te function y = f(x) at x. Formally,

2 Definition. Te derivative of te function y = f(x) at x is f(x + ) f(x) f (x) 0 or equivalently, Notes δx 0 δy δx. if Q approaces P from te rigt (left) ten > 0 ( < 0).. te notation does not indicate a ratio, it represents te limit of a ratio.. Limits In te above diagram, y = f(x), P is a fixed point, Q is a variable point, so x is a fixed number, and is a variable. Let Q P from eiter te left ( < 0) or rigt ( > 0), and consider te values of f(x + ) as 0. If 0 and > 0 ( < 0) write 0+ ( 0 ). Consider te limits lim 0 f(x + ), lim 0+ f(x + ). If a limit gives a definite value, te limit is said to exist. If it does not give a definite value, te limit does not exist. For te function wose grap is sown in te diagram, te tree numbers f(x), lim 0+ f(x + ), lim 0 f(x + ) all exist and are equal. In tis case y = f(x) is said to be continuous at x. If it appens tat te tree numbers are not all equal or some of tem do not exist, ten y = f(x) is discontinuous at x. Many of te functions tat we are familiar wit are continuous, for example all polynomials, sinx, cosx, expx. But, many, wile being continuous for most values of x also ave some points of discontinuity, for example rational functions (at points were te denominator is zero but te numerator is not), tanx, cosecx, secx, cotx. Example.. Let f(x) = x and x = ten lim 0 ( + ) = 0+ ( + ). In tis case, bot limits and te value of te function exist and all tree numbers are equal.. Let f(x) = and x = 0 ten f(0) is not defined. Also, if x = and > 0 ten as x 0, takes larger and larger values witout bound. We write f(x) as x 0+. Similarly, if < 0, takes negative values wit larger and larger magnitude. We write f(x) as x 0. In tis case, te value of te function is not defined and eiter te limits do not exist, or if one allows te possibility of ± as a limit, ten te limits are not equal. 3. Te function f(x) = x + x x

3 is not defined at x = 0 since division by zero is not allowed. It is not te same function as g(x) = x +. But for any value of x oter tan 0, we do ave f(x) = g(x) = x + and ence, since g(x) is continuous, lim f(x) = x 0 i.e. as x becomes closer and closer to 0, f(x) becomes closer and closer to. In tis case te value of te function does not exist, wile te limits exist and are equal. Now we cange notation and allow x to be a variable. Teorem.3 Properties of Limits Let c be a fixed number, and f(x) and g(x) functions of x. Ten provided te limits exist lim (f(x) + g(x)) f(x) + lim g(x) x c x c x c i.e. te limit of a sum equals te sum of te limits. lim (f(x)g(x)) f(x) lim g(x) x c x c x c i.e. te limit of a proct equals te proct of te limits. Te proof is beyond te scope of tis course. An important example of a limit is te following: Teorem.4 For θ measured in radians, as θ 0, we ave sin θ. θ Alternatively, sin θ lim =. θ 0 θ Proof: Once again, we cannot substitute θ = 0 as tis gives a division by zero. We use a geometric argument. Let te circle be of radius r. D θ C A B

4 Since te line CD is a tangent to te circle, angle ADC is a rigt angle. Now consider te tree areas: ADB < sector ADB < ADC. Te area of ADB is r sin θ, te area of te sector is θ π πr = r θ and te area of ADC is r tan θ. Hence, dividing by r gives r sin θ < r θ < r tan θ sin θ < θ < tan θ Te first inequality gives sin θ < θ wile te second gives cos θ < sin θ θ. Now let θ 0 ten cos θ and so sin θ is trapped between and a number tending to. θ Hence te result..3 Differentiation Te process of finding te derivative of a function is called differentiation. Having defined te derivative as te gradient of a curve, we now sow ow to use te definition to find derivatives algebraically. Example.5 Let y = x. Ten we ave f(x + ) f(x) (x + ) x 0 0 x + x + x x + (x + ) = x So, te derivative of x is x. Te next example generalises tis result to iger powers of x. Example.6 Let y = x n, wit n a positive integer. Ten f(x + ) f(x) (x + ) n x n 0 0 Expand (x + ) n using te binomial teorem only te first few terms will be needed: ( x n + nx n + n(n )xn + n(n )(n 6 )xn n x n ) 0

5 (nx n + n(n )xn + 6 ) n(n )(n )xn n 0 cancelling first te x n s and ten an from te numerator and denominator. Now, let 0 ten all but te first term becomes zero, so = nxn Note: Cancelling te s, eliminated te problem of division by 0 and effectively enabled us to put = 0 in te remaining formula. Example.7 Let y = sin x. Ten f(x + ) f(x) 0 0 sin (x + ) sin x Now, in te identity sin X sin Y = cos ( X+Y ) sin ( X Y ), let X = x +, Y = x ten cos ( x+ and from teorem.4, sin ( ) 0 0 ( ) sin ( ) cos (x + )sin ( ) (cos (x + ) ) lim 0 0 ) ( sin ( ) ) as 0, wic leaves us wit (cos (x + ) ) = cos x 0 Similarly, we may differentiate x n for positive integral n, and cos x (see example seets) and also te functions e x and ln x. We summarise a few important derivatives wic sould be committed to memory..3. Some Important Derivatives Function x n sin x cos x e x Derivative nx n cos x sin x e x ln x x

6 Teorem.8 Multiplication by a Scalar and a Sum of two functions Let u(x) and v(x) be functions of x wose derivatives are known and let k be a constant number.. Ten te derivative of te function k u(x), is given by d (k u) = k. Ten te derivative of teir sum u(x) + v(x), is given by d (u + v) = + dv Proof: Using te definition of derivative, let f(x) = u(x) + v(x) df f(x + ) f(x) u(x + ) + v(x + ) u(x) v(x) 0 0 u(x + ) u(x) v(x + ) v(x) u(x + ) u(x) + lim 0 v(x + ) v(x) = + dv A function f(x) may be te proct of two simpler functions, e.g. f(x) = (x + )e x is a proct of x + and e x. Ten f(x) may be differentiated using te proct rule. Teorem.9 Proct Rule Let f(x) = u(x)v(x), were u(x) and v(x) are functions of x. Ten d (u(x)v(x)) = v + udv Proof: First Metod. Let y(x) = u(x)v(x). As x is incremented to x + δx, let u, v, y cange to u + δu, v + δv, y + δy ten y(x + δx) = y + δy = (u + δu)(v + δv) and so i.e. δy = (u + δu)(v + δv) uv δy = uδv + vδu + δuδv Using te definition of derivative, δy δx 0 δx uδv + vδu + δuδv δx 0 δx

7 i.e. ( u δv δx 0 δx + v δu ) δx + δuδv δx now using te teorems about limits, ( u δv ) ( + lim v δu ) ( + lim δu δv ) δx 0 δx δx 0 δx δx 0 δx δv = u lim δx 0 δx + v lim δu δx 0 δx + lim δu δx 0 = u dv + v wic is te required formula, and completes te proof. lim δx 0 Second metod. Again, using te definition of derivative, df f(x + ) f(x) 0 u(x + )v(x + ) u(x)v(x) 0 u(x + )v(x + ) u(x + )v(x) + u(x + )v(x) u(x)v(x) 0 ( ) v(x + ) v(x) u(x + ) u(x) u(x + ) + v(x) 0 but te limit of a sum is te sum of te limits ( ) ( ) v(x + ) v(x) u(x + ) u(x) u(x + ) + lim v(x) 0 0 and letting 0, u(x + ) u(x), giving v(x + ) v(x) = u(x) lim 0 wic is te required formula. + v(x) lim 0 = u dv + v Example.0 Let y = x sin x. Let u = x and v = sin x. Ten δv δx u(x + ) u(x) = d (uv) = v + udv = (sin x)x + x (cos x) = x sin x + x cos x Example. Let y = xln x. We can let u = x and v = ln x. Ten = d (uv) = v + udv = (ln x) + x x = + ln x

8 Suppose we wis to differentiate te function y = sin (x ). Tis is a function of a function in te sense tat te output from te function x is te input for te sine function. Let u = x, ten y = sinu, and we now ave two simpler functions, eac of wic we can differentiate differentiate. Te following teorem enables us to differentiate te composite function y = sin (x ). A function like tis is called a function of a function. In te example, y is te function sin of te function x. Teorem. Cain Rule or Function of a Function Rule Let y = f(u) and u = g(x) ten te derivative of te function y = f(g(x)) is given by Proof: From te definition of derivative = δy δx 0 δx were δx represents an increment in x and δy representing te corresponding increment in y. Let te cange in u corresponding to an increment δx in x be δu ten and so Using te teorems on limits, δy δx = δy δu δu δx δy δx 0 δu ( δx 0 δu δx. ) ( δy lim δu δx 0 ) δu. δx = wic is te derivative of y wit respect to u multiplied by te derivative of u wit respect to x. Example.3 Differentiate y = sin (x ). Let u = x ten y = sinu and Hence, using te cain rule = = cos u, = x. = (cos u)x = x cos x Wit practice it is possible to use te cain rule witout explicitly making te substitution.

9 Example.4 Let y = e sin x. Let u = sin x, ten y = e u. Now, applying te cain rule: = = eu cos x = cos x e sin x Example.5 Let y = ln (x + 3). Let u = x + 3, ten y = ln u. Now, applying te cain rule: = = u = x + 3 Example.6 Let y = w = w for any function w of x. Now, applying te cain rule: = dw Teorem.7 Quotient Rule. Let dw = w dw = w y = u v = uv for any functions u and v of x. Let w = v and apply te proct rule: = w ( + udw = v + u ) dv v Tis is known as te quotient rule. = v u dv v Example.8 Let y = ex. So, using te quotient rule: x = v u dv v = xex e x x = ( x x dw ) e x Example.9 Let y = tan x = sin x. So, we using te quotient rule: cos x = v u dv cos x cos x sin x ( cos x) = v cos x = cos x = sec x Using te basic functions, and te rules for differentiating procts, quotients and function of a function, it is possible to differentiate a wide range of functions. Two last examples sould suffice: ex Example.0 Let y =. Tis is a quotient, so we use tat rule, but along te way ex sin x needs to be differentiated using te cain rule. = v u dv v = (sin x)(xex) (e x )(cos x) sin x = (xsin x cos x) e x sin x

10 Example. Let y = ( + cos x ) 6. Tis is a function of a function of a function: let u = +cos x ten y = u 6 may be differentiated wit respect to u. To differentiate u = +cos x we may use te cain rule again to differentiate u w.r.t x. So let v = x ten u = + cos v. Te formula for te derivative becomes = Note: tis is te proct of all te functions, differentiated wit respect to teir argument and dv dv multiplied togeter (justifying te term cain rule). Hence, = 6u5 ( sin v) x = 6( + cos x ) 5 ( x sin x ) = x sin x ( + cos x ) 5.4 Implicit Differentiation Let y = f(x) ten y is called an explicit function of x, e.g. y = x. Substituting a value of x gives a value of y. However, consider te example wic represents te equation of a circle. x + y =. In tis case substituting a value of x suc tat x + merely gives a value for y. To find y furter operations are required, including taking te square root. Here, y is not a function of x, since rearranging te expression to get y = ± x, we may ave two values for y, or value or even none ( a function must give exactly one value). Te general case is often written f(x, y) = 0 and is called an implicit relation between x and y. In order to differentiate an implicit expression, we proceed as follows. Example. In capter we obtained te trigonometric identity sin x = sin x cos x. Suppose we differentiate bot sides (te left is a function of a function, wile te rigt is a proct): cos x = (sin x)( sin x) + (cos x)(cos x) = (cos x sin x) and cos x sin x is te formula for cosx, so, differentiating an identity also gives an identity. We can also differentiate: d ( x + y ) = d ( ) d x + d y = 0 To obtain d (y ) we use te cain rule: let u = y, ten = = y.

11 Hence x + y = 0 and rearranging tis gives: = x y Note tat te rigt and side contains y, and tis may be obtained by solving te implicit relation for y, if required. Implicit differentiation is very useful in finding te derivatives of inverse functions..5 Example Let y = tan x, te inverse tan function (and not ). We can rearrange te formula so as tan x to eliminate te inverse function in favour of tan by taking te tangent of bot sides, to obtain x = tan y and ten differentiate bot sides: d ( x ) = d (tan y) = sec y Now, from te identity sec y = + tan y = + x, we get = + x wic is te formula for te derivative of tan x. We end tis section on differentiation by a summary of results. Some of tese ave been proved, or solved in examples. Oters are on exercise seets. Function x n sin x cos x tan x cot x sec x cosec x e x Derivative nx n cos x sin x sec x cosec x sec x tan x cosec x cot x e x ln x x sin x cos x tan x x x + x

12 Applications of te Derivative Te derivative as been defined as te gradient of a curve. In oter words, represents te rate of cange of y wit respect to x. As an example, if y represents distance, and x time, tis becomes te rate of cange of distance wit respect to time, i.e. speed. Note also, if we differentiate again, we measure te rate of cange of speed, i.e. acceleration. Suppose we ave a function f(x). We may interpret df = 0 as follows. f(x) Wen df = 0, te gradient of te grap is zero and te tangent to te curve is orizontal. A point at wic tis appens is called a stationary point or critical point. Tere are tree possibilities. Te diagram sows a local maximum and te gradient is canging from positive to negative as we move from left to rigt troug te stationary point. Similarly, we may draw a local minimum and in tis case te gradient canges from negative to positive as we move from left to rigt troug te stationary point. Suc points are called turning points. Te tird possibility is tat te sign of te gradient remains te same (eiter positive or negative) eiter side of te stationary point and tis is called a point of inflection. An example is provided by te curve of y = x 3 at te point x = 0, te gradient is zero tere but te gradient is positive everywere else.. Example Find te maximum value of te function y = + 3x x Solution: Differentiating, = 3 x = 0 wic gives x =.5. Substituting tis into te equation, we get y = 4.5. To see weter it is a maximum or a minimum, we may calculate te gradient eiter side of x =.5, so we dece tat 4.5 is te maximum. f (.4) = 0. > 0 f (.6) = 0. < 0

13 A function may ave several values of x were infinitely many) maxima and minima. = 0, for example, sin x as many (in fact Tere is an alternative way of classifying stationary points, wic involves finding te second derivative of te function and evaluating it at te stationary point. Teorem. Second derivative test. Let y = f(x) and let = 0 at te point x = x 0. Ten. if d y > 0 at x = x 0 ten te stationary point is a minimum.. if d y < 0 at x = x 0 ten te stationary point is a maximum. 3. if d y = 0 at x = x 0 ten te test fails and furter investigation is needed. Example. Find and classify te turning points of te function y = x 3 3x +. Solution: Differentiating, = 3x 3 = 0 so tere are two turning points at x = and x =. Using te second derivative test: d y = 6x If x = + ten d y = +6 wic is positive, so we ave a minimum. If x = ten d y = 6 wic is negative, so we ave a maximum. Let x = 0 ten y =. Tis gives sufficient information to sketc te curve.

14 3 More examples of maxima and minima can be found on te problem seets.