Derivative Drill on Chain Rule: NOTE: The first part of the worksheet only involves chain rule using trigonometric and exponential functions. There is a second part which incorporates inverse trig and log functions. Derivative Drill on Chain Rule Part 1: A. Differentiate the following: 1. f() = tan(57) 2. g() = tan (3 - ) 3. H() = tan(sin(4)) 4. y = e x 5. y = e sec 6. K() = ee 5 7. H(w) = cos( w)- cos(w) 8. g() = sec 4 () + sec4 9. y = esin(x) sin(ex) 1 csc r 10. y = 1 3r B. Perform the following differentiations: 1. Dt (sin(at)) 2. Da(sin(at)) 3. d 3y (e cos(4 3x)) dx 4. d 3y (e cos(4 3x)) dy C. Quotient Rule should only be applied when you are differentiating a function of t divided by another function of t. For example, you should use quotient rule to differentiate a function such as F(t) = 1 tan(3t) . However, if your function consists of a constant 1 tan(3t) divided by a function of t quotient rule is not only inappropriate but will likely take twice as long as should be necessary and frequently lead to errors. To differentiate a function such as y = 4 one should first rewrite y = 4(t + 3sin(5t)) -1 and then apply chain rule t 3sin(5t) to get y' = -4(t + 3sin(5t)) -2 (1+15cos(5t)). Use this technique to differentiate the following without using quotient rule. 1. y = 5 3 e 4x 2. g(t) = 8 e sin(x) 3. H(u) = 3 5 3 2 u 2 tan(u) 2 D. Let s(t) = 5 + 10sin(2t) be the position (cm) of a particle at time t (sec), where 0 < t < 1. 1. For what interval(s) of time will the particle be moving backwards? 2. The critical value of a function is any place where its derivative is either zero or undefined. Find all the critical values of the function s(t) on the interval [0, 1]. Describe the behavior of the partical at the critical values. 3. Find the acceleration function of the particle. For what values of t in [0, 1] will the particle's velocity be increasing? E. The concentration C (ppm or parts per million) of a chemical in a lake is given by the function C = C0 e where t > 0 is time (days), C0 is the chemical's concentration at t = 0, and k > 0 is a constant. kt 1. Determine the rate of change of the concentration with respect to time. Include units with your answer. 2. Is your answer to (a) always positive or always negative? Is more chemical being added to the lake or is the chemical being removed? 3. Larger values for the constant k indicate what about the chemical's rate of change? F. The force of attraction between two objects of mass m and M is given by F = gmM/r 2 where g is the gravitational constant and r is the distance between the two objects. If mass is being measured in kilograms, time in seconds, and distance r in meters, then the units of F are in Newtons (1 N = 1 (kgm)/sec2). 1. Determine the rate of change of F with respect to distance r. 2. What are the units of dF/dr? 3. Give a physical explanation of why F'(r) is always negative. 4. Suppose the distance r between the two objects changes with time; r = r(t). How fast will the force F change with respect to time? 5. What are the units of F'(t)? 6. Suppose r(t) = e-.004t. How fast is the force changing relative to time? What is the sign of F'(t)? Give a physical interpretation. G. Suppose that a population P is modeled by the differential equation dP/dt = 3.9P - .2P1.4 where P(t) gives the # of individuals P at time t (days). 1. What are the units of P'(t)? What does P'(t) represent? 2. Use the chain rule to determine P''(t). 3. What does P''(t) represent? What are its units? Derivative Drill on Chain Rule Part 1: Solutions: sec2 (3 ) 2 dy dy 3. H'() = 4cos(4) sec2(sin(4)) 4. 5. - e x - sec tan e sec dx dx sin w sin w 5 e 5 6. DK = 5 e e 7. H'(w) = + 2 w 2 cos w A. 1. f'() = 356sec2(57) 2. g'() = 8. g'() = 4sec4 () tan() + 43sec(4) tan(4) 9. y'= cos(x) e sin x sin e x e sin x cos e x e x 1 1 1 1 csc cot 1 3r 3csc 2 dy r r r r 10. 2 dr 1 3r B. 1. Dt (sin(at)) = acos(at) 2. Da(sin(at)) = tcos(at) d 3y 3. (e cos(4 3x)) = 3e3y sin(4 - 3x) dx d 3y 4. (e cos(4 3x)) = 3e3y cos(4 3x) dy C. 1. dy 20e 4x dx 3 e 4x 2 2. g'(x) = 4 cos x e sin x e sin x 3/ 2 5sec2 u 3 3. H'(u) = (2 u)2 (2 tan(u))2 D. 1. The particle moves backwards when the velocity v(t) = 20cos(2t) is negative. This occurs on the interval (.25, .75) 2. v(t) = 0 at times t = .25, .75. At these times the particle's motion is stopped. The direction of motion is switching from forwards to backwards at t = .25 and from backwards to forwards at t = .75. (Why?) 3. a(t) = -4002sin(2t). The velocity increases when a(t) > 0. This occurs for t in (.5, 1). E. 1. C'(t) = -kC0 e ppm/day 2. C'(t) is always negative. It is not possible to determine if the chemical is being added or removed from the lake. However, the concentration of the chemical is decreasing. (This could happen either by adding more water to the lake or by removing the chemical.) 3. Larger values for the constant k indicate what about the chemical's rate of change? The larger the value of k, the more negative the rate of change of the concentration. Thus, the larger the value of k, the more rapidly the concentration of the chemical is dropping. kt F. 1. F'(r) = 2gMm r3 2. The units are N/m. 3. As the objects move farther away from each other, the attractive force decreases; F is a decreasing function of r and hence F'(r) < 0. dF dF dr 2gMm dr 4. dt dr dt dt r3 5. The units are N/sec. 2gMm dF dF dr 2gMm dr .004e 0.004t .008gMme 0.008t 6. = 3 3 dt dr dt dt 0.004t r e F' > 0, indicating that F is an increasing function of time. G. 1. The units of P'(t) are #individuals/day. P'(t) represents the rate of change in the number of individuals in the population with respect to time. 2. d2P 2 3.9P'(t) 2.8P0.4P'(t) dt 3. The units of P"(t) are (#individuals)/(day)2. P"(t) is the rate of change in the population's rate of change with respect to time. Derivative Drill on Chain Rule Part 2: A. Differentiate the following functions: 1. F(t) = ln(tan-1(3t)) 2. g(t) = e arcsin(t) - 4sin-1(4et) - 4cos(ln(t)) 3. y = ln(x) tan-1(x) 4. H() = 3 ln 5. y = sin-1(ln(x)) + ln(3 - sec(x)) 6. F(x) = ln(tan(x)) + tan(ln(x)) + tan(x) ln(x) 7. y = ln(ln(x)) + tan-1(tan-1(x)) B. The volume V of a sphere of radius r (in cm) is given by the formula V = 1. Compute dV . Provide units. dr 4 3 r . 3 2. Now suppose that r = r(t); that is, suppose that r is changing over time. Measure t in sec. One might for instance think of the example of a snowball that is melting. Find a formula for dV . Include units. dt C. For each of the following pictures, write as a function of h. Then compute 1. 2. h 10 h 8 3. Now suppose that h is changing over time; h = h(t). Find a formula for Derivative Drill on Chain Rule Part 2: Solutions d . dt d . dh 1. F(t) = ln(tan-1(3t)) F'(t) = 2. g(t) = e g'(t) = arcsin(t) e arcsin t 1 t2 1 3 tan (3t) 1 9t2 - 4sin-1(4et) - 4cos(ln(t)) 16e t 1 16e2t 4 sin(ln(t)) t dy arctan(x) ln(x) dx x 1 x2 3. y = ln(x) tan-1(x) 4. H() = 1 3 ln H'() = 3 3 2 5. y = sin-1(ln(x)) + ln(3 - sec(x)) y' = 1 x 1 ln2 x sec x tan x 3 sec x 6. F(x) = ln(tan(x)) + tan(ln(x)) + tan(x) ln(x) F'(x) = 2 sec2 x sec ln x tan x ln x sec2 x tan x x x 7. y = ln(ln(x)) + tan-1(tan-1(x)) 1 1 y' = xln x 1 tan 1 x 1 2 2 1x B. The volume V of a sphere of radius r (in cm) is given by the formula V = 1. Compute dV dV . Provide units. 4r2 dr dr 4 3 r . 3 2. Now suppose that r = r(t); that is, suppose that r is changing over time. Measure t in sec. One might for instance think of the example of a snowball that is melting. Find a formula for dV dV dV dr dr 4r2 . Include units. , where the units are cm3/sec dt dr dt dt dt C. 1. = arcsin(h/10) (Note: This could also be done using arctangent or arccosine functions. The inverse sine function is the easiest to work with in this example. Why?) 2. = arctan(8/h) (Note: This could also be done using arcsine or arccosine functions. The inverse tangent function is the easiest to work with in this example. Why?) 3. d d dh dt dh dt 1 h2 10 1 100 , for the function in ex 1. d d dh 8 1 8 , for the function in ex 2. 2 h2 64 dt dh dt 64 h 1 2 h *********************************************************************************