Calculus 1 Lia Vas Finding and Using Derivative Te sortcuts We ave seen tat te formula f f(x+) f(x) (x) = lim 0 is manageable for relatively simple functions like a linear or quadratic. For more complex functions, finding te derivative using tis definition is not very effective. Because of tis, many sortcuts to finding derivative ave been introduced. Derivative of a constant function. If c is a constant and f(x) = c for every value of c, ten f(x+) f(x) = c c = 0. Tus, te derivative is zero. Alternatively, te same conclusion could be reaced by noting tat a orizontal line as te slope zero. Tus, f(x) = c f (x) = 0 Derivative of te power function. In Examples 6 of previous section, we ave seen tat te derivative of te line mx + b is m. Tus te derivative of x is 1. In Example 7 we ave seen tat te derivative of x is x. Let us demonstrate a more general formula wic will compute te derivative of x n for any positive integer n. Let f(x) = x n and let us find te formula for f (a) using te alternative formula f (a) = f(x) f(a) lim x a for derivative at a. For tis function te quotient f(x) f(a) becomes xn a n. To determine te limit of tis wen x a, we want to factor te numerator. Recall tat x n a n factors x a x a x a as x n a n = (x a)(x n 1 + x n a +... + xa n + a n 1 ) To convince yourself of tis formula, foil te rigt and side and obtain x n + x n 1 a +... + x a n + xa n 1 ax n 1 x n a... xa n 1 a n and note tat all te terms cancel except te first x n and te last one a n. Tus you ave x n a n. Tus, f x n a n (a) = x a lim x a = lim (x a)(x n 1 + x n a +... + xa n + a n 1 ) x a x a lim x a xn 1 + x n a +... + xa n + a n 1 = a n 1 + a n 1 +... + a n 1 + a n 1 = na n 1 Since f (a) = na n 1 we ave = Te Power Rule. f(x) = x n f (x) = nx n 1 Note tat tis formula confirms our calculation of derivative of x and x. Indeed wen n = 1, it produces te derivative of x = x 1 as 1x 1 1 = 1x 0 = 1 and wen n =, te derivative of x as x 1 = x 1 = x. Togeter wit te following tree rules, we sall be able to find derivative of any polynomial function witout using te definition of derivative. 1
Te Sum Rule. y = f(x) + g(x) y = f (x) + g (x) Te Difference Rule. y = f(x) g(x) y = f (x) g (x) Te Constant Multiple Rule. y = cf(x) y = cf (x) Tus, 1. Te derivative of te sum is te sum of te derivatives.. Te derivative of te difference is te difference of te derivatives. 3. To find te derivative of a constant multiple of te function, carry te constant and find te derivative of te function. Tese formulas old basically because te same rules can be applied to limits: limits are additive and constant factors out of tem. Tus, if y = f(x) + g(x), te sum rule olds since y = lim 0 f(x + ) + g(x + ) f(x) g(x) f(x + ) f(x) = lim 0 Te difference rule can be sown similarly. If y = cf(x), y = lim 0 cf(x + ) cf(x) = lim 0 c(f(x + ) f(x)) g(x + ) g(x) + = c lim 0 f(x + ) f(x) = f (x)+g (x) = cf (x) Tese rules enables you to find derivative of any polynomial function by differentiating term by term. Eac term is of te form ax n wic as derivative nax n 1 by power and constant multiple rules. Example 1. Find derivative of f(x) = x 3 + 1 4 x 5. Solution. f (x) = d dx x3 + d 1 dx 4 x d 5 = dx (3)x3 1 + 1 4 x 1 0 = 6x + 1x. General Power Rule. In practice problem 5c of te previous section, we ave seen tat te derivative of 1 = x x 1 is 1 = 1x. Note tat tis follows te pattern of te power rule: for n = 1, x te formula nx 1 produces exactly 1x. It can be sown tat te power rule olds for any real number n. Te General Power Rule. f(x) = x n f (x) = nx n 1 olds for every real n Tis rule enables us to find derivatives of functions wit negative or fractional powers. You need to make sure tat your function is written in te form x n before you apply te power rule. Te following algebra rules may be useful wen doing tat. 1 x n = x n n x = x 1/n
Example. Find derivatives of te following functions. (a) f(x) = x4 4 + 4 x 4 (b) f(x) = x 3 3 x Solution. (a) Before finding derivative, write bot terms of f(x) in ax n form: f(x) = 1 4 x4 +4x 4. Ten find derivative using te power rule for bot terms. f (x) = d 1 dx 4 x4 + d dx 4x 4 = 1 4 (4)x4 1 + 4( 4)x 4 1 = x 3 16x 5. If necessary, you can write your answer as f (x) = x 3 16. x 5 (b) Write bot terms of f(x) in ax n form first: f(x) = (x 3 ) 1/ 3x 1/ = x 3/ 3x 1/. Ten find derivative using te power rule for bot terms. f (x) = d dx x3/ d dx 3x 1/ = 3 x3/ 1 3( 1 )x 1/ 1 = 3 x1/ + 3 x 3/. If necessary, you can write your answer as f (x) = 3 x + 3. x 3 Te sortcuts. Having te differentiation formulas enables you to find te instantaneous rate of cange, te slope of te tangent line at a point, velocity or any oter rate of cange in an applied problem witout using te definition of te derivative. We revisit several problems of te previous section and present a sorter solution to tem. Example 3 Examples from previous section revisited. 4 Example 1 (b) Let f(x) = x +4. Find te instantaneous rate of cange of f(x) for x = 1. Ten find te equation of te tangent line to f(x) at x = 1. 4 Example (b) Assume tat te distance traveled by a moving object x second after te object started moving can be computed by f(x) = x + 4 feet. Determine te velocity of te object one second after it started moving. 4 Example 4. Find te derivative of f(x) = 1 at x =. x Solutions. For 4 1 (b), find te derivative f (x) = d dx x + d 4 = dx x1 1 + 0 = x. Ten plug x = 1 and obtain f (1) = (1) =. Tis computes te slope of te tangent line. Te equation of te tangent line can be obtained using te point-slope equation wit m = and (1, f(1)) = (1, 5). So, te tangent line is y 5 = (x 1) y = x + 3. For 4 (b) Find te velocity just as in 1(b) and obtain te velocity of feet per second. For 4 Example 4, use te power rule for f(x) = x 1 to find te derivative f (x) = 1x 1 1 = x = 1 x. Plug x = to obtain tat f () = 1 = 1 4. Example 4. Sow tat for te linear function f(x) = mx + b te average rate of cange from x = a to x = a + and te instantaneous rate of cange at x = a are equal. Solutions. Te average rate of cange is f(a+) f(a) m. Te instantaneous rate of cange at a is te limit of f(a+) f(a) = m(a+)+b (ma+b) = ma+m+b ma b = m = = m wic is also equal to m. Alternatively, you can find te instantaneous by noting tat te derivative f (x) = mx 1 1 + 0 = m. More on velocity. Using te previous example, you can relate te formula computing te velocity of an object wit distance s(t) at time t as v = ds wit te pre-calculus formula you used dt for velocity v = s now better expressed as v = s. Te first formula computes te instantaneous t t velocity wile te second formula computes te average velocity. As we ave seen in te previous 3
example, if te distance function is a linear function (tat is, if velocity is constant), te two formulas amount to te same ting. One sould keep in mind tat te velocity cannot be computed by te pre-calculus formula v = s wen s is a nonlinear function. t More on oter applications of derivative. Similar argument can be made for relation between several oter pysical quantities. For example, let us recall several oter equivalent examples: (1) te force is te quotient of cange in work and cange in displacement caused by te force, () te current produced by a movement of electric carge is te quotient of te carge and time, (3) te density of a piece of wire is te quotient of mass and te lengt of te wire. Te average velocity Te average force Te average current Te average density v = s ds, te velocity v = t dt F = W x, te force F = dw dx I = Q dq, te current I = t dt ρ = m dm, te density ρ = x dx Example 5. Te mass of a metal rod in kilograms depends on te lengt x measured in meters starting at te rod s end and can be computed by m(x) = 6 3 x. Determine te density of te rod 1 meter from te rod s end. Solution. Te density ρ at x = 1 can be computed as te value of derivative dm dx x=1 (recall tat tis is te same as m (1)). Note tat m(x) = 6x 1/3 so tat te derivative is m (x) = 6 1 3 x1/3 1 = x /3 = 3. Tus x m (1) = 3 1 = kilograms per meter. Some furter examples are given in practice problems below. Practice problems. 1. Find te derivative of te given functions. (a) y = x 5 3x 3 + 5x 9 (b) y = x 38 + 6 (c) y = x3 + x 3 (d) y = 4 x 1 3x 6. Find an equation of te line tangent to te curve at te indicated point. (a) f(x) = x + x at x =. (b) f(x) = x 3 + 3 x at x = 1. 3. Find te points on te given curve at wic te tangent as te given slope. Ten find te tangent lines at tose points. (a) f(x) = x 3 3x 5, m = 0. (b) f(x) = x 3 3 x +, m = 6. 4. A company determines tat its cost function is C(x) = 1000 + 35x.01x, 4
0 x 300, were x is te number of items produced and C(x) is te cost of producing x items in dollars. (a) Find te average rate of cange in cost wen x is canging from 100 to 150. (b) Find te instantaneous rate of cange in cost wen producing 00 items. 5. Assume tat te matematical model for te growt of a locust tree in its first century of life is given by (t) = 3 t, 0 t 100, were t is te age of te tree in years and (t) is te eigt of te tree in feet. Find (64) and (64) and interpret te meaning of your answers in a full sentence. 6. Te mass of bacteria culture at time t in ours, is approximated by N(t) = 4t 7/, in milligrams. (a) Find N(9) and N (9) and interpret te meaning of your answers in a full sentence. (b) Find ow fast te mass of bacteria increases 4 ours after te experiment started. 7. Te body mass index (BMI) is a number obtained as BMI = 703w were w is te weigt in pounds and is te eigt in inces. For a 15-lb female tat is now 65 inces tall but growing, calculate ow fast is BMI canging wit eac new inc. Explain te meaning of te answer. 8. If a stone is dropped from a building 150 feet tall, its eigt above te ground after t seconds is given by s(t) = 150 16t, in feet. Since s measures te distance from te ground, it is decreasing as time passes by so tat te value of te velocity is negative. Te absolute value of velocity is often referred to as speed. (a) Find te average speed of te stone between 1 and 3 seconds after it is dropped. (b) Find te speed.5 seconds after te stone is dropped. (c) Find te time te stone its te ground and te speed at te time of te impact. 9. A blood vessel can be considered to be a cylindrical tube of radius R and lengt l. Te velocity of te blood can be computed by te formula v = P 4ηl (R r ) were r measures te distance from te central axis and P and η are constants representing te pressure difference between ends of te tube and viscosity of te blood. Tis formula is known as te law of te laminar flow and reflects te fact tat te velocity is largest along te central axis (wen r = 0) and te smallest along te wall (wen r = R). Determine te formula computing te velocity gradient, te instantaneous rate of cange of velocity wit respect to r and comment on te sign of your answer. 10. A particle moves on a line away from its initial position so tat after t ours it is s(t) = t 1 miles from its initial position. (a) Find te velocity of te particle 5 ours after it started moving. (b) Find te time wen te velocity is 30 miles per our. 11. Te mass of a bacteria culture t ours after te start of experiment, is modeled by N(t) = 3t 5/, in milligrams. (a) Determine te mass 16 ours after experiment started. (b) Determine ow fast te mass of bacteria increases 9 ours after te experiment started. (c) Determine te time wen te mass is 300 mg. 5
Solutions. 1. (a) y = (5)x 5 1 3(3)x 3 1 + 5(1)x 1 1 0 = 10x 4 9x + 5 (b) y = 38x 38 1 + 0 = 38x 37 (c) y = x3 + x 3 = 1 x3 + x 3/ y = 1 (3)x3 1 + 3 x3/ 1 = 3 x + 3 x1/ or y = 3x + 3 x. (d) y = 4 1 = 4x 1 x 3x 6 3 x 6 y = 4( )x 1 1 3 ( 6)x 6 1 = 8x 3 + x 7 or y = 8 + x 3 x 7. (a) To use te point-slope equation, you need to compute te y-value of point wit x = and te slope f (). f() = + = 1 + 1 =. To find te derivative, note tat f(x) = x 1 + 1x f (x) = ( 1)x 1 1 + 1 x1 1 = + 1. Tus f () = + 1 = 1 + 1 = 0. So, te tangent line is x y = 0(x ) y =. (b) f(1) = 1 3 + 3 1 = 1 + 1 =. So, te function passes (1,). f(x) = x 3 + 3 x = x 3/ + x /3 f (x) = 3 x1/ + 3 x 1/3. At x = 1, f (1) = 3 11/ + 3 1 1/3 = 3 + = 9+4 = 13. 3 6 6 Tus, te tangent line is y = 13(x 1) y = 13x 1. 6 6 6 3. (a) f(x) = x 3 3x 5 f (x) = 3x 3 1 3()x 1 0 = x 6x. Te problem is asking you to find te values of x for wic f (x) = 0 x 6x = 0 x(x 6) = 0 x = 0 and x = 6. Tus, te curve as orizontal tangent at points 0 and 6. f(0) = 0 3 3(0) 5 = 5 so te tangent is y + 5 = 0(x 0) y = 5. Wen x = 6, f(6) = 108 5 = 103. Te tangent is y 103 = 0(x 6) y = 103. (b) f(x) = x 3 3 x + f (x) = 3x 3 1 3 x 1 +0 = 3x 3x. Te problem is asking you to find te values of x for wic f (x) = 6 3x 3x = 6 3x 3x 6 = 0 3(x )(x+1) x = and x = 1. Tus, te curve as tangent wit slope 6 at points and -1. Since f() = 8 3 4 + = 4 te tangent at x = is y 4 = 6(x ) y = 6x 8. Wen x = 1, f( 1) = 1 3 + = 1 so te tangent is y + 1 = 6(x + 1) y = 6x + 11. 4. (a) Wen production canges from 100 to 150 items produced, te cost increased at an average rate of C(150) C(100) = 605 4400 = 3.5 dollars per item produced. (b) C (x) = 0 + 35x 1 1 150 100 50 0.01()x 1 = 35 0.0x. Wen producing 00 items, te cost is increasing at a rate C (00) = 35 0.0(00) = 31 dollars per item produced. 5. (t) = 3t 1/. (64) = 4. 64 years after it starts growing, te tree is 4 feet tall. (t) = 3 1 t1/ 1 = 3 t 1/ = 3. t (64) = 3 =.1875 0.19. 64 years after it starts growing, te tree 16 is growing at te rate of.19 feet per year. 6. (a) N(t) = 4t 7/, N(9) = 8748 mg = te mass of bacteria 9 ours after. N (t) = 4 7 t7/ 1 = 14t 5/. N (9) = 340 mg per our. Tus, after 9 ours, te mass is increasing at te rate of 340 mg per our. (b) 4 ours after, te mass of bacteria is increasing at te rate of N (4) = 14(4) 5/ = 448 mg per our. 7. For w = 15, BMI() = 703(15) = 87875 BMI () = 87875( ) 1 = 175750. Wen 3 = 65, te value of te derivative is 175750.64. Tus, te BMI is decreasing by.64 per 65 3 inc. Te negative sign indicate tat for a fixed weigt, te BMI decreases wen te eigt increases. 6
8. (a) Te average velocity between 1 and 3 seconds after te stone is dropped is s(3) s(1) = 3 1 6 134 = 64. Tus te average speed is 64 feet per second. (b) s (t) = 3t s (.5) = 80 so te speed.5 seconds after te stone is dropped is 80 feet per second. (c) Te stone its te ground wen s(t) = 150 16t = 0 150 = 16t t = ± 150 ±3.06. Since we are 16 interested in positive value of time, we conclude tat te stone its te ground 0.06 seconds after it is dropped. Te velocity at te time of te impact is s (3.06) = 97.984. Tus, te speed is about 98 feet per second. 9. v = P 4ηl (R r ) dv = P (0 dr 4ηl r 1 ) = P decreases as r increases. 4ηl rp r =. Te sign is negative since te velocity ηl 10. (a) s(t) = t 1 s (t) = 4t s (5) = 0 so tat te velocity 5 ours after is 0 miles per our. (b) s (t) = 4t = 30 t = 30 = 7.5. Tus, te velocity is 30 miles per our 7.5 ours 4 after. 11. (a) N(t) = 3t 5/ N(16) =307 mg. (b) N (t) = 15 t3/ N (9) = 15 93/ = 0.5 mg per our. (c) N(t) = 3t 5/ = 300 t 5/ = 100 t = 100 /5 6.31 ours. Tus, te mass is 300 mg 6.31 ours after te experiment started. 7