Higher Derivatives. Differentiable Functions

Transcription

1 Calculus 1 Lia Vas Higer Derivatives. Differentiable Functions Te second derivative. Te derivative itself can be considered as a function. Te instantaneous rate of cange of tis function is te second derivative. Tus, te second derivative evaluated at a point computes te slope of te tangent line to te grap of te first derivative. Te second derivative f (x) is te derivative of te first derivative f (x) Notation: f d (x), 2 f(x), y, d2 y dx 2 dx 2 Evaluated at x = a : f d (a), 2 f(x) dx 2 x=a, d2 y dx 2 x=a Example 1. Find te second derivative of te following functions. (a) f(x) = 3x 2 + 5x 6 (b) f(x) = 7 x 2 Solutions. (a) f(x) = 3x 2 + 5x 6 f (x) = 6x + 5 f (x) = 6x = 6. (b) f(x) = 7 x 2 = 7x 2 f (x) = 14x 3 f (x) = 14( 3)x 3 1 = 42x 4 = 42 x 4 Applications. If s(t) represents te distance an object moved in time t, we ave seen tat te derivative s (t) represents te velocity at time t. Te second derivative is te acceleration since it calculates te rate at wic te velocity is canging. velocity acceleration v(t) = s (t) = ds dt a(t) = v (t) = dv = dt s (t) = d2 s dt 2 Te sign of te acceleration determines if te object is speeding up or slowing down as follows. velocity positive ten velocity negative ten acceleration positive object speeds up acceleration negative object slows down acceleration positive object slows down acceleration negative object speeds up Tink of acceleration and velocity as two vectors. If te two vectors ave te same sign, te acceleration adds to te increase in velocity so tat te object speeds up. If te two vectors ave te opposite sign te acceleration causes te velocity to decrease so tat te object slows down. Anoter way to interpret te conclusions in te table above is to relate te speed to te sign of acceleration and velocity. Recall tat speed is te absolute value of te velocity. Tus, 1

2 velocity > 0 ten acceleration > 0 speed positive and increasing object speeds up acceleration < 0 speed positive and decreasing object slows down velocity < 0 ten acceleration > 0 speed positive and decreasing object slows down acceleration < 0 speed positive and increasing object speeds up Example 2. Consider te problem wit a stone being dropped from a 150 feet building from practice problem 8 in te previous section. Te eigt above te ground after t seconds is given by s(t) = t 2 feet. Determine te acceleration and discuss te sign of te velocity and acceleration. Solution. s(t) = t 2 v(t) = s (t) = 32t a(t) = s (t) = 32. Te velocity is negative because te distance from te ground is decreasing as time passes. Te negative acceleration indicates tat tis negative velocity is becoming more negative in time, tus te stone speeds up te more time passes by. Te value of te answer means tat te object is accelerating 32 feet per second every second. Tus, te speed is increasing by 32 feet per second every second. Example 3. Consider te object wose distance traveled (in meters) is computed as a function of time (in seconds) described by te formula s(t) = t 3 7t t. Find te formulas for velocity and acceleration, grap te tree functions and determine wen te object speeds up and wen it slows down. Solution. s(t) = t 3 7t t v(t) = s (t) = 3t 2 14t + 13 a(t) = s (t) = 6t 14. Te tree graps are given on te figure on te rigt. From te grap we can determine tat te object speeds up on intervals were v(t) and a(t) ave te same sign: approximately 1.3 < t < 2.3 and t > 3.4. Te object slows down on intervals were v(t) and a(t) ave te opposite sign: approximately 0 < t < 1.3 and 2.3 < t < 3.4. Higer Derivatives. Continuing differentiating te derivative, one obtains te iger derivatives: te second derivative as te derivative of te first, te tird derivative as te derivative of te second and so on. Te tird derivative is usually denoted by f (x). For te derivatives iger tan tree, f (n) is used to denote te n-t derivative. So, for example, te fourt derivative is written as f (4) (x) rater tan f (x). Example 4. Find te fourt derivative of te function f(x) = x + x Solution. f(x) = x + x = x 1/2 + x f (x) = 1 2 x1/ x = 1 2 x 1/2 + 2x f (x) = x 1/ x 0 = 1 4 x 3/2 + 2 f (x) = 1 4 f (4) (x) = x 5/2 1 = x 7/ x 3/ = 3 8 x 5/2 2

3 f(a+) f(a) Differentiable Functions. Recall tat te derivative at x = a is te limit lim 0 or, f(x) f(a) equivalently lim x a. If eiter one (ten necessarily bot) of tese limits exist, f(x) is said to x a be differentiable at x = a. Tus, if you can find te derivative of a function (eiter by definition or using te differentiation formulas) and if f (x) is defined at x = a, ten te function is differentiable at a. Recall tat a function is continuous at x = a if limit of f(x) wen x a exists and it is equal to f(a). Tus, f(a) = lim x a f(x) or, equivalently, lim x a f(x) f(a) = 0. If a function f(x) is differentiable at a, ten lim f(x) f(a) = lim x a x a f(x) f(a) f(x) f(a) (x a) = lim x a x a lim x a x a = f (a) 0 = 0 x a and so te function is continuous. Tus, if a function is differentiable, it is continuous. Te contrapositive 1 of tis last claim is stating tat if a function as a discontinuity at a ten it is not differentiable at a. Even wen continuous, f(x) may fail to be differentiable at a if te left and rigt limits of f(a+) f(a) are different. In tis case, te slope of te tangent on te left and te slope of te tangent on te rigt side of a are different and f(x) is said to ave a corner or a sarp turn at x = a. Te last scenario of f(x) failing to be differentiable at a is wen eiter left, rigt (or bot) limits of f(a+) f(a) exist but are not finite. In tis case, f(x) is said to ave a vertical tangent. Tus, f(x) can fail to be differentiable at x = a in any of te following cases. 1. f(x) is not continuous at a. 2. f(x) as a corner at a. 3. f(x) as a vertical tangent at a. Example 5. Discuss te differentiability of te following { functions. x x 0 (a) f(x) = x (b) f(x) = x + 1 x < 0 (c) f(x) = x (d) f(x) = 3 x Solution. (a) Te derivative f (x) of f(x) at any x is 1 (eiter by using te power rule or finding f(x+) f(x) x+ x it using te definition as lim 0 = lim 0 = lim 0 = 1. Since f (x) = 1 is defined at every x, f(x) = x is differentiable for every x. (b) For x > 0, f(x) = x so f (x) = 1 and tus f(x) is differentiable for every x > 0. Similarly, for x < 0, f(x) = x + 1 so tat f (x) = 1 and tus f(x) is differentiable for every x < 0 as well. At x = 0, te function is not continuous since te left limit wen x 0 is 1 and te rigt limit wen x 0 + is 0. Since f(x) is not { continuous at 0, it is not differentiable at 0 as well. x x 0 (c) Recall tat x = x x < 0. For x > 0, we ave seen tat f (x) = 1 so f(x) is differentiable for every x > 0. Similarly, for x < 0, f (x) = 1 so f(x) is differentiable for every x < 0 as well. If it exists, te derivative f f(0+) f(0) 0+ 0 (0) at x = 0 is equal to lim 0 = lim 0 = lim 0. Since te value of depends on te fact if is positive or negative, we consider tese cases separately: 1 A contrapositive of te implication of te form p q is te statement tat not q not p. For example, a contrapositive of te statement if a polygon is a triangle, it as tree sides is if a polygon does not ave tree sides, it is not a triangle. 3

4 1. wen > 0, = and so lim 0 + = lim 0 + = wen < 0, = and so lim 0 = lim 0 = 1. Tus, te left and te rigt limit of f(0+) f(0) are not equal and so f (0) does not exist. Tus f(x) is not differentiable at 0. Looking at te grap of x one can notice it as a corner. (d) f(x) = 3 x = x 1/3 f (x) = 1 3 x 2/3 = Tis derivative is defined for every value x2 of x except 0. So, f(x) is differentiable for every x 0. f (x) is not defined at 0 so f(x) is not differentiable at 0. Te grap of f(x) reveals a vertical tangent at x = 0. If a function is given by a grap, it is differentiable at a point if it as a (non-vertical) tangent at (bot sides of) te point. If tere is a corner, discontinuity or a vertical tangent, it is not differentiable. Example 6. Discuss te differentiability of te function given by te grap on te rigt. Solutions. Te function is differentiable at every point different from -1 and 1 since tere is a well defined tangent to te grap for all x ±1. At x = 1 te function as a break so it is not continuous and tus also not differentiable. At x = 1 te function is not differentiable since tere is a corner in te grap. Practice problems. 1. Find te first and te second derivative of te following functions. (a) f(x) = x x 2 (b) f(x) = x x 2 2. Find te first five derivatives of te following functions. (a) f(x) = 2x 5 3x 3 + 5x 9 (b) f(x) = 2 x + x 2 3. An arrow as been sot in te air and its eigt above te ground is described by te formula s(t) = 24t 4.9t 2 were t is in seconds and s is in meters. 4

5 (a) Determine te acceleration, grap te eigt, velocity and acceleration on te same plot and determine wen te arrow speeds up and wen it slows down by discussing te sign of te velocity and acceleration. (b) Determine te time te arrow is at te igest distance from te ground. (c) Determine te time te arrow falls down to te ground and its speed at te time of te impact. 4. Discuss te differentiability of te following functions. (a) f(x) = x (b) f(x) = 5 3 x 2 (c) f(x) = 2 3x 1/5 (d) Te function given by te grap on te rigt. Solutions. 1. (a) f(x) = x3 + 4 = 1 2 x 2 2 x3 + 4x 2 f (x) = 3 2 x2 8x 3 f (x) = 3x + 24x 4 = 3x + 24 x 4 (b) f(x) = x x 2 = x 3/2 + x 2/3 f (x) = 3 2 x1/ x 1/3 f (x) = 3 4 x 1/2 2 9 x 4/3 2. (a) f(x) = 2x 5 3x 3 + 5x 9 f (x) = 10x 4 9x f (x) = 40x 3 18x f (x) = 120x 2 18 f (4) (x) = 240x f (5) (x) = 240. (b) f(x) = 2 x + x 2 = 2x x f (x) = 2x f (x) = 4x 3 f (x) = 12x 4 f (4) (x) = 48x 5 f (5) (x) = 240x (a) s(t) = 24t 4.9t 2 v(t) = s (t) = t a(t) = 9.8. Grap te tree functions on te same plot. Note tat te eigt increases and te velocity is positive wile te arrows goes up wic appens approximately in te first 2.5 seconds. After tat te eigt decreases and velocity is negative until te arrow its te ground about 5 seconds after it as been sot. Te acceleration is constant and negative. Tus, te arrow slows down in te first 2.5 seconds (velocity and acceleration ave te opposite signs) and it speeds up between 2.5 and 5 seconds (velocity and acceleration ave te same sign). 5

6 (b) Te arrow is at te igest distance from te ground exactly wen velocity is zero. Tus, v(t) = t = 0 24 = 9.8t t = seconds. We can see tat our estimate from 9.8 part (a) is close to tis answer. (c) Te arrow falls down to te ground wen te eigt is at zero. s(t) = 24t 4.9t 2 = 0 t(24 4.9t) = 0 t = 0 or t 4.9. At time t = 0 it as been sot in te air and at te time t = 4.9 seconds it falls down to te ground. We can see tat our estimate from part (a) is close to tis answer. Te velocity at te time of te impact is v(4.9) = (4.9) 24 meters per second. Tus, te speed is about 24 meters per second. 4. (a) f(x) = x f (x) = 2x. Te derivative is defined at every x-value so f(x) is differentiable for every x. (b) f(x) = 5 3 x 2 = 5x 2/3 f (x) = 10 3 x 1/3 = Tis function is defined for every value of x x except x = 0. Graping f(x), you can notice tat it as a corner (and a vertical tangent) at x = 0 so it is not differentiable at 0. Tus, f(x) is differentiable for every x 0. (c) f(x) = 2 3x 1/5 f (x) = 3 5 x 4/5 = Tis function is defined for every value of x x4 except x = 0. Graping f(x), you can notice tat it as a vertical tangent at x = 0 so it is not differentiable at 0. Tus, f(x) is differentiable for every x 0. (d) Te function is differentiable at every point different from -2, 0 and 1 since tere is a well defined tangent to te grap for all x 2, 0, 1. At x = 2 and x = 0 te function is not differentiable since it is not continuous (a jump at -2 and a ole at 0). At x = 1 te function is not differentiable since tere is a corner in te grap. 6