unit 2

rate of change

"dependent" per "independent"

find f(π/6) if f(x) = x / secx

(6√3 - π) / 12

find the average rate of change over the interval 2<=x<=30. x: 0, 2, 7, 30 f(x): 3, -2, 5, 7

(7 - -2)/(30 - 2) = 9/28

average rate of change: the following quotients express the average rate of change of a function over an interval.

(f(a+h) - f(a))/((a+h) - a) or (f(x) - f(a))/ (x - a) this is also the slope of the secant line

d/dxcotx =

-csc^2x

d/dxcscx =

-cscxcotx

find the average rate of change of each function on the given interval 1. w(x) = lnx; 1 <= x <= 7 2. s(t) = -t^2 - t + 4; [1,5] 3. find the derivative of y = 2x^2 + 3x - 1 by using the definition of the derivative. 4. for the function h(t), h is the temperature of the oven in Fahrenheit, and t is the time measured in minutes. a. explain the Maning of the equation h(15) = 420 b. explain the meaning of the equation h'(43) = -11

1. 0.324 2. -7 ft/sec 3. y' = 4x + 3 4. a. After 15 minutes, the temperature of the oven is 420 degrees F b. At the 43th minute, the oven temperature is decreasing by 11 degrees F/min

find the derivative of each function (some of which are at a given x-value) 1. y = 5 - cscx 2. h(x) = 2xtanx 3. r = sinθ/θ 4. g(x) = cotx/x 5. f(x) = 1/2cosx 6. y = 5xsecx 7. f(x) = 3tanx at x = 2π/3 8. f(x) = 2secx at x = π/4 9. f(x) = xcotx at x = π/6

1. y' = cscxcotx 2. h'(x) = 2tanx + 2xsec^2x 3. r' = (cosθ(θ) - sinθ) / θ^2 4. g'(x) = (-xcsc^2x - cotx) / x^2 5. f'(x) = sinx/2cos^2x 6. y' = 5secx + 5xsecxtanx 7. 12 8. 2√2 9. √3 - 2π/3

sec =

1/cos

estimate the derivative at the given x-value by using a calculator 10. f(x) = sin^(2)(x/5) at x = 1.8 11. f(x) = cot(x^2) / 2 at x = -1 12. f(x) = 3sec(e^x) at x = 2.5

10. 0.132 11. 1.412 12. -15.923

find f'(a) for each function at the given value of a. 11. f(x) = x^4; find f'(-1) 12. f(x) = √x; find f'(16) 13. f(x) = 1/x^4; find f'(2) 14. f(x) = 1/(^3√x); find f'(27)

11. -4 12. 1/8 13. -1/8 14. -1/243

find the equation of the tangent line of the function at the given x-value 12. f(x) = -2x^3 + 3x at x = -1 13. f(x) = 4sinx - 2 at x = π 14. find the equation for the normal line of y = 1/2x^2 + 3/4x - 4 at x = -3 15. if f(x) = 3sinx - 2e^x find f'(0)

12. y + 1 = -3(x + 1) 13. y + 2 = -4(x - π) 14. y + 7/9 = 4/9(x + 3) 15. 1

find the equations of both the normal line and the tangent line 13. y = secx at x = π 14. y= tanx at x = π/3

13. tangent: y = -1 normal: ? possibly x = π but prolly not 14. tangent: y - √3 = 4(x - π/3) normal: y - √3 = -1/4(x - π/3)

find the equation of the tangent line at the given x-value 13. f(x) = sinx/cosx at x = π/3 14. g(x) = -2x/e^x at x = 0

13. y - √3 = 4(x - (π/3) 14. y = -2x

find the equations of the tangent and normal lines of each function at the given value of x. 14. f(x) = 3√x + 4 at x = 4 15. y = (x^2)/2 +3/2x - 2 at x = 8 16. f(x) = -x^3 + 2x^2 - 2 at x = 2

14. tangent: y - 10 = 3/4(x - 4); normal: y - 10 = -4/3(x - 4) 15. tangent: y - 42 = 19/2(x - 8); normal: y - 42 = -2/19(x - 8) 16. tangent: y + 2 = -4(x - 2); normal: y + 2 = 1/4(x - 2)

16. if f(x) = sin(3x^2 - 2); find f'(7) 17. if f(x) = csc(3x) at x = 2 18. use the table below to estimate the value of d'(120). t seconds: 2, 13, 60, 180, 500 d(t) feet: 10, 81, 412, 808, 2105 19. is the function differentiable at x = 2? f(x) = {3x - 3x^2 - 5, x < 2; 7 - 9x, x >= 2 20. what values of a and b would make the function differentiable at x = 4? f(x) = {a√x + bx^2 - 1, x < 4; 16/x + bx, x>= 4

16. 260.246 17. -36.899 18. 3.3ft/sec 19. yes 20. a = 47/11; b = -13/44

when do the two functions listed have parallel tangent lines? 18. f(x) = x^2 and g(x) = x^5 19. f(x) = √x and g(x) = x^3

18. x = 0 and x = ^3√2/5 19. x = 0.488

the table below shows values of two differentiable functions f and g, as well as their derivatives. x: 2, -5 f(x): 4, 3 f'(x): -2, 4 g(x): -1, -2 g'(x): 2, 5 4. h(x) = 3f(x)g(x); find h'(2) 5. r(x) = ((f(x)/2) + 2)(3-g(x)); find r'(-5)

4. 30 5. -7.5

find the derivative of each function 5. f(x) = 4 - (1/(2x^2)) 6. g(x) = 3√x - (6/x^2) + 5π^3 7. h(x) = 4e^x - 2cosx 8. s(t) = t^2sin(t) 9. d(t) = 3√(t)lnt 10. y = 4/x - secx 11. h(x) = (2-x)/(x+2)

5. f'(x) = 1/x^3 6. g'(x) = (3/2√x) + (12/x^3) 7. h'(x) = 4e^x + 2sinx 8. s'(t) = 2tsin(t) + (t^2)(cos(t)) 9. d'(t) = (3lnt/2√x) + (3√t/t) 10. y' = -4/x^2 - secxtanx 11. h'(x) = -4/((x+2)^2)

use the following table to find the average rate of change on the given interval. t (minutes): 0, 3, 4, 12, 13 s(t) (feet): -2, 4, -7, 5, 10 6. [3,13] 7. 0<=x<=12 8. [3,4]

6. 0.600 ft/min 7. 7/12 ft/min 8. -11 ft/min

find derivative: 7. f(x) = x/√x; find f'(7) 8. f(x) = ^3√(x)*(x^3). find f'(8)

7. f'(7) = 1/2√7 ^ this is the slope of the tangent line at x = 7 8. f'(8) = 1280/3

derivatives of exponential functions: d/dx a^x = d/dx e^x =

a^(x)lna e^(x)lne = e^x

d/dxsinx =

cosx

derivative

equation for slope of tangent line (slope of function at exact point)

the power rule:

f(x) = x^n f'(x) = nx^(n-1)

function: f(x) = x^2 f(x) = x^3 f(x) = x^4 f(x) = x^5 find the function's derivative:

function's derivative: f'(x) = 2x f'(x) = 3x^2 f'(x) = 4x^3 f'(x) = 5x^4

normal lines: a normal lines...

goes through the same point the tangent line does, but it is perpendicular to the tangent line

product rule

h(x) = f*g h'(x) = d/dx[f(x)g(x)] = (f(x)*d/dx[g(x)]) + (g(x)*d/dx[f(x)]) *you don't need the product rule if it doesn't have an "x"

quotient rule

h(x) = f/g h'(x) = d/dx [f(x)/g(x)] = ((g(x)*d/dx[f(x)]) - (f(x)*d/dx[g(x)])) / g(x)^2

if it starts with c...

it is negative

instantaneous rate of change: the following limits express the instantaneous rate of change of a function at x = a.

lim ((f(a+h) - f(a))/h) h→0 or lim ((f(x) - f(a))/(x-a)) x→a this is also the slope of the tangent line at x = a

each limit represents the instantaneous rate of change of a function. identify the original function, and the x-value of the instantaneous rate of change. 21 through 28

look at page 4 of the packet

use the tables to estimate the value of the derivative at the given point. indicate units of measure. 14 through 19 a and b

look at the practice in 2.3 estimating derivatives

identify any x-values of the function that are not continuous and/or not differentiable 1 through 3

look at the practice page of 2.4 differentiability and continuity

identify points where the function below is not continuous and/or not differentiable

look on page 1 of the 2.4 differentiability and continuity notes

d/dxsecx =

secxtanx

definition of the derivative

this limit gives an expression that calculates the instantaneous rate of change (slope of the tangent line) of f(x) at any given x-value f'(x) = lim ((f(x+h) - f(x))/h) h→0

5. if we know h(5) = -2, and the derivative of h is given by h'(x) = (x^3 - 2) / x, write an equation for the line tangent to the graph of h at x = 5.

y + 2 = (123/5)(x - 5)

find derivative: 1. y = x^37 2. y = x^9

y' = 37x^36 y' = 9x^8

d/dxcosx =

-sinx

derivative rules: constant: (d/dx)*c = constant multiple: (d/dx)*cu = sum/difference (d/dx)(u+/-v) =

0 ex: y = e^2; y' = 0 c*(du/dx) ex: y = (x^5)/3, y = 1/3x^5, y' = 5/3x^4; or y = 2x^3, y' = 3*2x = 6x du/dx +/- du/dx

ln1= ln0= e^0= e^(lna)= lne^a=

0 undefined 1 a a

Estimating the Derivative with a CALCULATOR: 1. if f(x) = sin √x, find f'(2) 2. If f(x) = ln(1/(5-x)), find f'(1.3) 3. Write the equation of the line tangent to y = √x/(x^3 + 1) at x = 1

1. 0.055 2. 0.270 3. y - 0.707 = -0.177(x - 1)

find the average rate of change of each function on the given interval. use the appropriate units if necessary. 1. f(x) = x^2 - 2; [-1,3] 2. A(t) = 2^t; [2,4]; t represents years, A represents dollars 3. h(m) = tan(m) + 4; [π/4, 3π/4]; h represents hair, m represents months 4. a(x) = lnx on the interval 2<=x<=7 5. f(x) = cosx on the interval -1<=x<=0

1. 2 2. $6/year 3. -1.273 hair/months 4. 0.251 5. 0.460

estimate the derivative at the given point by using a calculator. 1. f(x) = x√(2 - x); find f'(-10) 2. f(x) = sec(5x); find f'(2) 3. f(x) = ln(√x); find f'(1) 4. f(x) = e^(x/3); find f'(4) 5. f(x) = tan(sinx); find f'(1.3) 6. f(x) = 2^(ln(x)); find f'(2) 7. the model f(t) = x/cosx measures the height of bird in meters where t is seconds. find f'(2). 8. the model f(t) = sin^2(t) measures the depth of a submarine measured in feet where t is minutes. find f'(12.5) 9. the model f(t) = √(x) - (1/(x - 1)) measures the number of stocks sold where t is days. find f'(12)

1. 4.907 2. -3.864 3. 0.500 4. 1.265 5. -1.010 6. 0.560 7. 8.098 m/sec 8. -0.132 ft/min 9. 0.153 stock/day

find the derivative using limits. if the equation is given as y =, use Leibniz Notation: dy/dx. if the equation is given as f(x) =, use Lagrange Notation: f'(x). 1. f(x) = 7 - 6x 2. y = 5x^2 - x 3. y = √(5x + 2) 4. f(x) = 1/(x - 2)

1. f'(x) = -6 2. dy/dx = 10x - 1 3. dy/dx = 5/(2√(5x + 2)) 4. f'(x) = -1/((x - 2)^2)

find the derivative of each function: 1. f(x) = 2lnx 2. f(x) = 5^x 3. f(x) = 9cosx 4. f(x) = 5e^x 5. f(x) = 8lnx - 4cosx + e 6. f(x) = 15sinx = 3e^x 7. f(x) = log 2 x 8. f(x) = log 7 x + cosx 9. f(x) = 3^x + 3x + x^3 10. f(x) = 3sinx

1. f'(x) = 2/x 2. f'(x) = 5^(x)ln5 3. f'(x) = -9sinx 4. f'(x) = 5e^x 5. f'(x) = 8/x + 4sinx 6. f'(x) = 15cosx - 3e^x 7. f'(x) = 1/xln2 8. f'(x) = 1/xln7 - sinx 9. f'(x) = 3^(x)ln3 + 3 + 3x^2 10. f'(x) = 3cosx

find the derivative of each function: 1. f(x) = 2sinx + 5e^x 2. f(x) = 3^x - 4cosx 3. f(x) = log 2 x - sinx

1. f'(x) = 2cosx + 5e^x 2. f'(x) = 3^(x)ln3 + 4sinx 3. f'(x) = 1/(xln2) - cosx

find the derivative of each function: 1. f(x) = (2x-3)cosx 2. g(x) = 2x^(3)e^x 3. h(x) = 4√(x)lnx 4. f(x) = (4-5x)cosx 5. g(x) = 6lnxsinx 6. h(x) = 2e^x(x^2 + x) 7. f(x) = 8sinxcosx 8. g(x) = (3/x)lnx 9. h(x) = 2x^(5)cosx 10. f(x) = e^(x)sinx

1. f'(x) = 2sinx + (2x - 3)cosx 2. g'(x) = 6x^(2)e^x + 2x^(3)e^x 3. h'(x) = (2lnx/√x) + (4√x/x) 4. f'(x) = 5cosx - (4 - 5x)sinx 5. g'(x) = (6/x)(sinx) + 6lnxcosx 6. h'(x) = 2e^(x)(x^2 + 3x + 1) 7. f'(x) = 8cos^2x - 8sin^2x 8. g'(x) = (-3lnx + 3)/x^2 9. h'(x) = 10x^(4)cosx - 2x^(5)sinx 10. f'(x) = e^(x)(sinx + cosx)

find the derivative of each function. 1. f(x) = 2x^3 - 4x + 5 2. g(x) = 5x^-2 - 1/2x^4 3. y = 2e^4 - 3x 4. y = πx^2 - π 5. y = 3x^2 - 1/6x^2 6. h(x) = (x^6)/3 + 6x^(2/3) - 4x^(1/2) + 2 7. f(x) = 1/√x + 3/5x 8. f(x) = √x + 3*(^3√x) + 2 9. f(x) = 3x^7 - 4x^3 + 5x + 7 10. y = 4√x + e

1. f'(x) = 6x^2 - 4 2. g'(x) = -10/x - 2x^3 3. y' = -3 4. y' = 2πx 5. y' = 6x(1/(3x^3)) 6. h'(x) = 2x^5 + 4/^3√x - 2/√x 7. f'(x) = -1/2√x^3 + -3/5x^2 8. f'(x) = 1/2√x + 1/(^3√x^2) 9. f'(x) = 21x^6 - 12x^2 + 5 10. y' = 2/√x

find the derivative of each function: 1. f(x) = 8xsinx 2. g(x) = 2e^(x)*(√x) 3. h(x) = ((1/x) + 1)(2x^2 - 5)

1. f'(x) = 8sinx+8xcosx 2. g'(x) = e^(x)*((2x+1)/√x) 3. h'(x) = 5/(x^2) + 4x + 2

identify the function we are working with. then identify the x-value for the instantaneous rate of change (slope of the tangent line at a point). 1. lim ((5ln(2/(4+h)) - 5ln(1/2))/h) h→0 2. lim ((sinx - 1)/(x - π/2)) x→π/2

1. function: f(x) = 5ln(2/x); instantaneous rate at x = 4 2. function: f(x) = sinx; instantaneous rate at x = π/2

find the derivative of each function: 1. h(x) = 4x-1 / 3x+2 2. g(x) = sinx / x 3. h(x) = (x^3 = 3x^2 - x) / 2x 4. h(x) = 4x/lnx 5. (3x^4 - 2x^2 - 3√x) / x 6. g(x) = 2x^5 / cosx 7. e^x / 4sinx 8. 2x+4 / 3x+2 9. (x^3 + 3x^2 - x) / x^2

1. h'(x) = 11 / (3x+2)^2 2. g'(x) = (xcosx - sinx) / x^2 3. h'(x) = x + 1 4. h'(x) = (4lnx - 4)/(lnx)^2 5. f'(x) = (3x^4 - 2x^2 - 3√x) / x 6. g'(x) = 2x^5 / cosx 7. f'(x) = e^x / 4sinx 8. f'(x) = -8x / (3x + 2)^2 9. g'(x) = 1 + 1/x^2

find the derivative of each function: 1. y = 2x^2 - 5/x + 6 2. y = 8√x - (x^6)/3 + 2pi^5

1. y' = 4x - 5/x^2 2. y' = 4/√x - 2x^5

find the derivative of each function: 1. y = 2x^2 / 3x+1 2. g(x) = 3e^x / 2x 3. h(x) = sinx / 2x^(2)-5 4. h(x) = 3x+1 / 2x^2

1. y' = 6x^(2)+4x / (3x+1)^2 2. g'(x) = 6e^(x)(x-1) / 4x^2 3. h'(x) = (cosx2x^(2)-5 - 4xsinx) / (2x^(2) - 5)^2 4. h'(x) = (-3/2x^2) - (1/x^3)

find dy/dx: 1. y = x^7 2. y = x 3. y = x^π 4. y = 1/x^5 5. y = 1/^4√x 6. y = ^9√x^4 7. y = 3^√x 8. y = x^e 9. y = x/^3√x 10. y = x^2(^6√x^5)

1. y' = 7x^6 2. y' = 1 3. y' = πx(π-1) 4. y' = -5/x^6 5. y' = -1/(4*(^4√x^5)) 6. y' = 4/(9*(^9√x^5)) 7. y' = 1/(3*(^3√x^2)) 8. y' = ex^(e-1) 9. y' = 2/(3*(^3√x)) 10. y' = 17/(6*(^6√x^11))

estimate the derivative with a calculator of g(x) = csc^(2)4x at x = 2

1.202

csc =

1/sin

derivatives of logarithmic functions d/dx log a x = d/dx lnx = d/dx log e x =

1/x * 1/lna 1/x * 1/lne = 1/x

use the table to find the value of the derivatives of each function 10. x:7, f(x):-5, f'(x):3, g(x):2, g'(x):-3 a. h(t) = 5f(x) / g(x); find h'(7) b. m(x) = g(x)+2 / 3f(x) 11. t:-4, a(t):2, a'(t):-3, b(t):-4, b'(t):1 a. f(t) = -b(t) / 3a(t): find f'(-4) b. g(t) = 1-a(t) / 2b(t)+3; find g'(-4) 12. x:1, d(x):-4, d'(x):-2, h(x):4, h'(x):3 a. g(x) = d(x)/2h(x); find g'(1) b. f(x) = 2-(d(x)/2) / (6 - h(x)); find f'(1)

10. a. -45/4 b. 1/25 11. a. 5/6 b. -13/25 12. a. 1/8 b. 7/2

for each function, write the equation of the tangent line at the given value of x. 10. f(x) = ln2x / 4x at x = 1 11. f(x) = cos(tan(x)) at x = 2 12. f(x) = x^4 / √x at x = 3 13. f(x) = x^2*sin(1/x) at x = 7

10. y - 0.173 = 0.077(x - 1) 11. y + 0.576 = 4.719(x - 2) 12. y - 46.765 = 54.560(x - 3) 13. y - 6.976 = 1.003 (x - 7)

use the table to find the value of the derivatives of each function. 11. x:7, f(x):-5, f'(x): 3, g(x): 2, g'(x): -3. a. h(t) = f(x)g(x); find h'(7) b. m(x) = 5f(x)g(x); find m'(7) c. s(x) = (3f(x) - 1)(g(x) + 2); find s'(7) 12. x:-4, a(t):2, a'(t):-3, b(t):-4, b'(t):1 a. f(t) = a(t)b(t); find f'(-4) b. g(t) = -3a(t)b(t); find g'(-4) c. h(t) = (1 - a(t))(3b(t) + 2); find h'(-4) 13. x:1, d(x):-3, d'(x):-2, h(x):4, h'(x):3 a. a(t) = d(x)h(x); find a'(1) b. b(x) = -d(x)h(x); find b'(1) c. c(x) = (2 - (d(x)/2))(6 - h(x)); find c'(1)

11. a. 21 b. 105 c. 84 12. a. 14 b. -42 c. -33 13. a. -17 b. 17 c.-17/2

find the value of the derivative at the given point 11. if f(x) = 3x - 6cosx, find f'(π/2) 12. if f(x) = √(x) - 2lnx, find f'(4) 13. if f(x) = 4e^x + 5sinx, find f'(0) 14. if f(x) = 2cosx + e^x, find f'(π)

11. 9 12. -1/4 13. 9 14. e^π

find the x-value(s) where the function has a horizontal tangent 11. f(x) = (x^3)/3 + 4x^2 + 12x - 13 12. f(x) = (x^4)/2 + x^3 + (x^2)/2 + 7 13. f(x) = (x^4)/4 - (10x^3)/3 + 21/x + 6/5

11. x = -6, x= -2 12. x = 0, x = -1/2, x = -1 13. x = 0, x = 7, x = 3

find the equation of the tangent line at the given x-value 14. f(x) = 8sinxcosx at x = π/3 15. g(x) = -2xe^x at x = 0

14. y - 2√3 = -4(x - π/3) 15. y = -2x

find the instantaneous rate of change of each function at the given x-value. use the form lim ((f(a+h) - f(a))/h) h→0 15. f(x) = x^2 - x at x = -1 16. f(x) = √x at x = 5 17. f(x) = 1/x at x = 2

15. -3 16. 1/(2√5) 17. -1/4

find the equation of the tangent line of each function at the given value of x. 15. y = x^3 at x = -2 16. f(x) = ^4√x^3 at x = 1 17. f(x) = 1/x^4 at x = 2

15. y - 8 = 12(x + 2) 16. y - 1 = 3/4(x - 1) 17. y - (1/16) = -1/8(x - 2)

find the equation of the tangent line at the given x-value 15. f(x) = 3cosx + x at x = π/2 16. f(x) = 4e^x - 3sinx + x^2 at x = 0

15. y - π/2 = -2(x - π/2) 16. y - 4 = x

are the functions differentiable at the given value of x? 17. at x = 5. f(x) = {2x - 8/5x^2 + 10, x<=5; 50 - 14x, x>5 18. at x = 9. f(x) = {30/√x - x^2, x<9; x^2 - 5x - 107, x>=9 19. at x = 3. f(x) = {5x^2 - 2x + 1, x<=3; 3x^2 + 2x + 6, x>3

17. yes 18. no 19. no

find the instantaneous rate of change of each function at the given x-value. use the form lim ((f(x) - f(a))/(x-a)) 18. f(x) = 2x^2 + 3x - 4 at x = -3 19. f(x) = √(3x) at x = 7 20. f(x) = 1/(5x) at x = -2

18. -9 19. 3/(2√21) 20. -1/20

what values of a and b would make the function differentiable at the given value of x? 21. at x = 2. f(x) = {ax^4 + x + 4, x<2; bx - 5, x>=2 22. at x = 1. f(x) = {a/x^2 + x^3 - 2, x<=1; x^2 + bx + 1, x>1

21. a = 3/16, b = 7 22. a = 4/3, b = -5/3

f(x) = x^2 f'(x) = f'(1) = f'(2) = f'(-2) = 1. if f'(x) = (5/x) - x, find f'(2) and explain the meaning 2. if f(x) represents how many meters you have run and x represents the minutes, describe in full sentences the following: f(8) = 1,500 f'(3) = 161

2x 2 4 -4 1. f'(2) = 1/2; the slope of the tangent line of f(x) at x = 2 is 1/2 2. after 8 minutes, I ran 1,500 meters at the 3rd minute, I was running 161 meters per minute

find the instantaneous rate of change of f(x) = x - x^2 at x = -1 do both ways

3 for h→0 3 for x→-1

find the derivative using the definition of the derivative (limits). 3. f(x) = 2x^2 - 7x + 1 4. y = 1/(x^2)

3. f'(x) = 4x - 7 4. dy/dx = -2/(x^3)

rewrite, differentiate, and simplify 3. function: y = 1/x 4. function: y = 1/x^4 5. function: y = √x 6. function y = ^7√x^3

3. rewrite: y = x^-1 differentiate: dy/dx = -1x^-2 simplify: dy/dx = -1/x^2 4. rewrite: y = x^-4 differentiate: y' = -4x^-5 simplify: y' = -4/x^5 5. rewrite: y = x^(1/2) differentiate: y' = 1/2x^(-1/2) simplify: y' = 1/(2√x) 6. rewrite: y = x^(3/7) differentiate: y' = 3/7x^(-4/7) simplify: y' = 3/(7*^7√(x^4))

the table below shows values of two differentiable functions f and g, as well as their derivatives x: 2 f(x): 4 f'(x): -2 g(x): -1 g'(x) = 2 5. h(x) = f(x)/3g(x); find h'(2) 6. r(x) = -(g(x) / (1-f(x))); find r'(2)

5. -2 6. 1/3

for each problem, use the information given to identify the meaning of the two equations in the context of the problem. write in full sentences. 5. C is the number of championships Sully has won while coaching basketball. t is the number of years since 2002 for the function C(t). C(12) = 3 and C'(12) = 0.4 6. d is the distance (in miles) from home when you walk to school. h is the number of hours since 7:00 am for the function d(h). d(0.5) = 1.2 and d'(0.5) = -11 7. W is the number of cartoon shows Mr. Kelly watches every week. x is the number of children Mr. Kelly has for the function W(x). W(7) = 25 and W'(7) = 3 8. g is the number of gray hairs on Mr. Brust's head. x is the number of students in his 4th grade period. g(26) = 501 and g'(15) = 130

5. In 2014, Sully had won 3 basketball championships. At year 12 of coaching basketball, Sully is winning 0.4 championships per year. 6. At 7:30 am, I am 1.2 miles from home. At 7:30 am, I am walking 11 miles per hour back home. 7. If Mr. Kelly had 7 kids, he watches 25 cartoons each week. If he had 7 kids, the rate of watching cartoons increases by 3 per week. 8. With 26 kids in Mr. Brust's 4th period, he has 501 gray hairs. With 15 kids in his 4th period, his gray hair count is increasing by 130 gray hairs per kid.

for each problem, create an equation of the tangent line of f at the given point. leave in point-slope. 9. f(7) = 5 and f'(7) = -2 10. f(-2) = 3 and f'(-2) = 4 11. f(x) = 3x^2 + 2x; f'(x) = 6x + 2; x = -2 12. f(x) = 10√(6x + 1); f'(x) = 30/√(6x + 1); x = 4 13. f(x) = cos2x; f'(x) = -2sin2x; x = π/4 14. f(x) = tanx; f'(x) = sec^2x; x = π/3

9. y - 5 = -2(x - 7) 10. y - 3 = 4(x + 2) 11. y - 8 = -10(x + 2) 12. y - 50 = 6(x - 4) 13. y = -2(x - π/4) 14. y - √3 = 4(x - π/3)

estimating the derivative with a calculator:

MATH [8]: nDeriv(

what values of a and b would make this function f(x) = [x^2 - ax + 2, x<3; x+b, x>=3 differentiable at x = 3?

a = 5 b = -7 solve for b by setting equations equal to each other b will still be an equation then set derivatives equal to each other and solve for a plug at into b to get real b

derivatives of cosx and sinx

d/dx cosx = -sinx d/dx sinx = cosx

estimating the derivative from tables: the function must be _______________ to estimate a derivative. this just means, the graph is ___________ and _________. 1. x (hours): 0, 2, 4, 7, 11; f(x) (miles): -2, 3, 10, 1, -3; using the table, estimate f'(3) 2. x (seconds): 10, 50, 80, 120, 150; w(x) (gallons per second): 950, 850, 700, 500, 150; using the table, estimate w'(100)

differentiable continuous smooth 1. 7/2 miles/hr 2. -5 gallons/second^2

if it is not continuous (the two equations do not equal the same number when set equal to each other), then it is not...

differentiable (set derivatives equal to eachother)

the derivative fails to exists where the function has a

discontinuity (hole, jump, or vertical asymptote) corner or cusp (where the line squishes into basically one line and dead flowers for lines as they extend from the squish) vertical tangent (part of the line is completely vertical, so its tangent line is vertical too)

Find the value of the derivative at the given point. 4. if f(x) = 3lnx + e^x. find f'(5)

f'(5) = 3/5 + e^5

notation for the derivative: Lagrange: Leibniz:

f'(x) dy/dx = the derivative of y with respect to x

find f'(x) if f(x) = log 4 x - 4lnx

f'(x) = 1/xln4 - 4/x

find f'(x) if f(x) = 2^x + 3e^x

f'(x) = 2^(x)ln2 + 3e^x

f'(x) if f(x) = 2sinx - 5cosx

f'(x) = 2cosx + 5sinx

3. find the x-values of any horizontal tangent lines of f(x) = 4x^2 + 7x - 13

f'(x) = 8x + 7 x = -7/8

parallel tangent lines: 9. let f(x) = x^4 and g(x) = x^3. at what value(s) of x do the graphs of f and g have parallel tangent lines

f'(x) = g'(x) 4x^3 = 3x^2 4x^3 - 3x^2 = 0 x^2(4x-3) = 0 x^2 = 0 4x-3 = 0 x = 0 and x = 3/4

if f(x) = 3cosx + sinx / 2, find f'(π)

f'(π) = -1/2

find the equation of a normal line of f(x) = x^3 - 4x^2 + x + 3 at x = 3

f(3) = -3 f'(x) = 3x^2 - 8x + 1 f'(3) = 4 y + 3 = -1/4(x - 3)

find the average rate of change of f(x) = ln 3x over the interval 1<=x<=4

f(4) - f(1) / (4 - 1) = (ln(3*4) - ln(3*1))/3 = 0.462

average rate of change on the interval [a,b] is represented by...

f(b) - f(a) / b - a

this derivative is an expression that calculates the...

instantaneous rate of change (slope of the tangent line) of a function at any given x-value. in other words, it gives us the slope of the function at a point

use the following graph to find the average rate of change on the given interval. 9, 10, and 11

look at page 3 in the packet

the graphs of f and g are given below. for each function, find the average rate of change on the given interval 12, 13, and 14

look at page 3 in the packet for compositions: f(g(x)), find the g(x) first and then use those numbers to plug into f(x) and find those points for addition h(x) = f(x) + g(x): solve for h(x) for both numbers on the interval by adding each f(x) and g(x) for each number

parallel = perpendicular =

same slope opposite reciprocal slope

d/dxtanx =

sec^2x

differentiability

the derivative exists for each point in the domain. the graph must be a smooth line or curve for the derivative to exist. in other words, the graph looks like a line if you zoom in. zoom in = looks like linear line

equation of the tangent line:

the line tangent to the curve of f(x) at x = a can be represented in point slope form: y - f(a) = f'(a)(x - a)

horizontal tangent lines: when does a function have a horizontal tangent line?

the slope of a horizontal tangent line is zero. to find where a function has a horizontal tangent line, we set the derivative equal to zero. generally are min/max

true or false: differentiability implies continuity continuity implies differentiability

true false

6. the graph of f'(x), the derivative of f, is shown at the right. if f(2) = 7, write an equation of the line tangent to the graph of f at (2,7)

y - 7 = 3(x - 2) (2,3) is the point on the graph f'(x)

find the derivative of y = sinxtanx

y' = sinx + sinxsec^2x

is the function f(x) = {5x^2 + 3x + 2, x<-1; -7x - 3, x>=-1 differentiable at x = -1?

yes