Derivatives of Trigonometric Functions
- dydx=2x−secxtanx
- dydx=2xcotx−x2csc2x
- dydx=xsecxtanx−secxx2
- dydx=(1−sinx)(1−sinx)−cosx(x+cosx)
- dydx=2csc2x(1+cotx)2
-
y=−x
-
y=x+2−3π2
-
y=−x
- d2ydx2=3cosx−xsinx
- d2ydx2=12sinx
- d2ydx2=2cscx(csc2x+cot2x)
- x=(2n+1)π4, where n is an integer
- (π4,1),(3π4,−1)
- a=0,b=3
- y′=5cos(x), increasing on (0,π2),(3π2,5π2), and (7π2,12)
- d3ydx3=3sinx
- d4ydx4=5cosx
- d3ydx3=720x7−5tan(x)sec3(x)−tan3(x)sec(x)
The Chain Rule
- dydx=18u2⋅7=18(7x−4)2⋅7
- dydx=−sinu⋅−18=−sin(−x8)⋅−18
- dydx=8x−242√4u+3=4x−12√4x2−24x+3
-
- f(u)=u3,u=3x2+1
- dydx=18x(3x2+1)2
-
- f(u)=u7,u=x7+7x
- dydx=7(x7+7x)6⋅(17−7x2)
-
- f(u)=cscu,u=πx+1
- dydx=−πcsc(πx+1)⋅cot(πx+1)
-
- f(u)=−6u−3,u=sinx
- dydx=18sin−4x⋅cosx
- dydx=4(5−2x)3
- dydx=6(2x3−x2+6x+1)2(3x2−x+3)
- dydx=−3(tanx+sinx)−4⋅(sec2x+cosx)
- dydx=−7cos(cos7x)⋅sin7x
- dydx=−12cot2(4x+1)⋅csc2(4x+1)
- 10
- −18
- –4
- –12
- 1034
- y=−12x
- x=±√6
-
- −200343 m/s
- 6002401m/s2
- The train is slowing down since velocity and acceleration have opposite signs
-
- C′(x)=0.0003x2−0.04x+3
- dCdt=100⋅(0.0003x2−0.04x+3)
- Approximately $90,300 per week
-
- dSdt=−8πr2(t+1)3
- The volume is decreasing at a rate of −π36ft3/min.
- ≈2.3 ft/hr
Derivatives of Inverse Functions
-
- (f−1)′(1)≈2
-
- (f−1)′(1)≈−1/√3
-
- 6
- x=f−1(y)=(y+32)1/3
- 16
-
- 1
- x=f−1(y)=sin−1y
- 1
- 15
- 13
- 1
-
- 4
- y=4x
-
- −113
- y=−113x+1813
- 2x√1−x4
- −1√1−x2
- 3(1+tan−1x)21+x2
- −1(1+x2)(tan−1x)2
- x(5−x2)√4−x2
- –1
- 12
- 110
-
- v(t)=11+t2
- a(t)=−2t(1+t2)2
- v(2)=0.2,v(4)=0.06,v(6)=0.03;a(2)=−0.16,a(4)=−0.028,a(6)=−0.0088
- The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.
- –0.0168 radians per foot
-
- dθdx=10100+x2−401600+x2
- 18325,9340,424745,0
- 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.
- −5412905,−3500,−19829945,−91360
- 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
- dydx=−2xy
- dydx=x3y−y2x
- dydx=y−y2√x+4√x+4−x
- dydx=y2cos(xy)2y−sin(xy)−xycosxy
- dydx=−3x2y−y3x3+3xy2
y=−12x+2
y=1π+12x−3π+38π+12
y=0
-
- y=−x+2
- (3,−1)
-
- (±√7,0)
- Slope is −2 at both intercepts
- They are parallel since the slope is the same at both intercepts.
- y=−x+1
-
- –0.5926
- 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.
- dydx=−8
- dydx=−2.67
- y′=−1√1−x2
Derivatives of Exponential and Logarithmic Functions
- f′(x)=2xex+x2ex
- f′(x)=ex3lnx(3x2lnx+x2)
- f′(x)=4(ex+e−x)2
- f′(x)=24x+2⋅ln2+8x
- f′(x)=πxπ−1⋅πx+xπ⋅πxlnπ
- f′(x)=52(5x−7)
- f′(x)=tanxln10
- f′(x)=2x⋅ln2⋅log37x2−4+2x⋅2xln7ln3
- dydx=(sin2x)4x[4⋅ln(sin2x)+8x⋅cot2x]
- dydx=xlog2x⋅2lnxxln2
- dydx=xcotx⋅[−csc2x⋅lnx+cotxx]
- dydx=x−1/2(x2+3)2/3(3x−4)4⋅[−12x+4x3(x2+3)+123x−4]
y=−15+5ln5x+(5+15+5ln5)
-
- x=e≈2.718
- y′>0 on (e,∞), and y′<0 on (0,e)
-
- P=500,000(1.05)t individuals
- P′(t)=24395⋅(1.05)t individuals per year
- 39,737 individuals per year
-
- 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.
- 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.
- p=35741(1.045)t
-
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