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Techniques for Differentiation: Get Stronger Answer Key

Derivatives of Trigonometric Functions

  1. dydx=2xsecxtanx
  2. dydx=2xcotxx2csc2x
  3. dydx=xsecxtanxsecxx2
  4. dydx=(1sinx)(1sinx)cosx(x+cosx)
  5. dydx=2csc2x(1+cotx)2
  6. y=x

    The graph shows negative sin(x) and the straight line T(x) with slope −1 and y intercept 0.

  7. y=x+23π2

    The graph shows the cosine function shifted up one and has the straight line T(x) with slope 1 and y intercept (2 – 3π)/2.

  8. y=x

    The graph shows the function as starting at (−1, 3), decreasing to the origin, continuing to slowly decrease to about (1, −0.5), at which point it decreases very quickly.

  9. d2ydx2=3cosxxsinx
  10. d2ydx2=12sinx
  11. d2ydx2=2cscx(csc2x+cot2x)
  12. x=(2n+1)π4, where n is an integer
  13. (π4,1),(3π4,1)
  14. a=0,b=3
  15. y=5cos(x), increasing on (0,π2),(3π2,5π2), and (7π2,12)
  16. d3ydx3=3sinx
  17. d4ydx4=5cosx
  18. d3ydx3=720x75tan(x)sec3(x)tan3(x)sec(x)

The Chain Rule

  1. dydx=18u27=18(7x4)27
  2. dydx=sinu18=sin(x8)18
  3. dydx=8x2424u+3=4x124x224x+3
    1. f(u)=u3,u=3x2+1
    2. dydx=18x(3x2+1)2
    1. f(u)=u7,u=x7+7x
    2. dydx=7(x7+7x)6(177x2)
    1. f(u)=cscu,u=πx+1
    2. dydx=πcsc(πx+1)cot(πx+1)
    1. f(u)=6u3,u=sinx
    2. dydx=18sin4xcosx
  4. dydx=4(52x)3
  5. dydx=6(2x3x2+6x+1)2(3x2x+3)
  6. dydx=3(tanx+sinx)4(sec2x+cosx)
  7. dydx=7cos(cos7x)sin7x
  8. dydx=12cot2(4x+1)csc2(4x+1)
  9. 10
  10. 18
  11. 4
  12. 12
  13. 1034
  14. y=12x
  15. x=±6
    1. 200343 m/s
    2. 6002401m/s2
    3. The train is slowing down since velocity and acceleration have opposite signs
    1. C(x)=0.0003x20.04x+3
    2. dCdt=100(0.0003x20.04x+3)
    3. Approximately $90,300 per week
    1. dSdt=8πr2(t+1)3
    2. The volume is decreasing at a rate of π36ft3/min.
  16. 2.3 ft/hr

Derivatives of Inverse Functions

    1. A curved line starting at (−3, 0) and passing through (−2, 1) and (1, 2). There is another curved line that is symmetric with this about the line x = y. That is, it starts at (0, −3) and passes through (1, −2) and (2, 1).
    2. (f1)(1)2
    1. A quarter circle starting at (0, 4) and ending at (4, 0).
    2. (f1)(1)1/3
    1. 6
    2. x=f1(y)=(y+32)1/3
    3. 16
    1. 1
    2. x=f1(y)=sin1y
    3. 1
  1. 15
  2. 13
  3. 1
    1. 4
    2. y=4x
    1. 113
    2. y=113x+1813
  4. 2x1x4
  5. 11x2
  6. 3(1+tan1x)21+x2
  7. 1(1+x2)(tan1x)2
  8. x(5x2)4x2
  9. 1
  10. 12
  11. 110
    1. v(t)=11+t2
    2. a(t)=2t(1+t2)2
    3. v(2)=0.2,v(4)=0.06,v(6)=0.03;a(2)=0.16,a(4)=0.028,a(6)=0.0088
    4. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.
  12. 0.0168 radians per foot
    1. dθdx=10100+x2401600+x2
    2. 18325,9340,424745,0
    3. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening.
    4. 5412905,3500,19829945,91360
    5. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should sit for maximizing the viewing angle is 20 feet

Implicit Differentiation

  1. dydx=2xy
  2. dydx=x3yy2x
  3. dydx=yy2x+4x+4x
  4. dydx=y2cos(xy)2ysin(xy)xycosxy
  5. dydx=3x2yy3x3+3xy2
  6. The graph has a crescent in each of the four quadrants. There is a straight line marked T(x) with slope −1/2 and y intercept 2. y=12x+2
  7. The graph has two curves, one in the first quadrant and one in the fourth quadrant. They are symmetric about the x axis. The curve in the first quadrant goes from (0.3, 5) to (1.5, 3.5) to (5, 4). There is a straight line marked T(x) with slope 1/(π + 12) and y intercept −(3π + 38)/(π + 12). y=1π+12x3π+38π+12
  8. The graph starts in the third quadrant near (−5, 0), remains near 0 until x = −4, at which point it decreases until it reaches near (0, −5). There is an asymptote at x = 0. The graph begins again near (0, 5) decreases to (1, 0) and then increases a little bit before decreasing to be near (5, 0). There is a straight line marked T(x) that coincides with y = 0. y=0
    1. y=x+2
    2. (3,1)
    1. (±7,0)
    2. Slope is 2 at both intercepts
    3. They are parallel since the slope is the same at both intercepts.
  9. y=x+1
    1. 0.5926
    2. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor.
  10. dydx=8
  11. dydx=2.67
  12. y=11x2

Derivatives of Exponential and Logarithmic Functions

  1. f(x)=2xex+x2ex
  2. f(x)=ex3lnx(3x2lnx+x2)
  3. f(x)=4(ex+ex)2
  4. f(x)=24x+2ln2+8x
  5. f(x)=πxπ1πx+xππxlnπ
  6. f(x)=52(5x7)
  7. f(x)=tanxln10
  8. f(x)=2xln2log37x24+2x2xln7ln3
  9. dydx=(sin2x)4x[4ln(sin2x)+8xcot2x]
  10. dydx=xlog2x2lnxxln2
  11. dydx=xcotx[csc2xlnx+cotxx]
  12. dydx=x1/2(x2+3)2/3(3x4)4[12x+4x3(x2+3)+123x4]
  13. The function starts at (−3, 0), decreases slightly and then increases through the origin and increases to (1.25, 10). There is a straight line marked T(x) with slope −1/(5 + 5 ln 5) and y intercept 5 + 1/(5 + 5 ln 5). y=15+5ln5x+(5+15+5ln5)
    1. x=e2.718
    2. y>0 on (e,), and y<0 on (0,e)
    1. P=500,000(1.05)t individuals
    2. P(t)=24395(1.05)t individuals per year
    3. 39,737 individuals per year
    1. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1963 there were approximately 723 cases of the disease in New York City.
    2. At the beginning of 1960 the number of cases of the disease was decreasing at rate of 4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of 0.2808 thousand per year.
  14. p=35741(1.045)t
  15.  
  16. Years since 1790 P
    0 69.25
    10 107.5
    20 167.0
    30 259.4
    40 402.8
    50 625.5
    60 971.4
    70 1508.5
  17.