Section 1
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-csc(x)+C
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Cards (177)
Section 1
(50 cards)
-csc(x)+C
ln(x)+C
Global Definition of a Derivative
Mean Value Theorem
The instantaneous rate of change will equal the mean rate of change somewhere in the interval. Or, the tangent line will be parallel to the secant line.
f'(x)-g'(x)
Horizontal Asymptote
Mean Value Theorem for integrals or the average value of a functions
f'(g(x))g'(x)
f'(x)+g'(x)
sec(x)tan(x)
Fundamental Theorem of Calculus #2
Extreme Value Theorem
If f is continuous on [a,b] then f has an absolute maximum and an absolute minimum on [a,b]. The global extrema occur at critical points in the interval or at endpoints of the interval.
1
Critical Number
If f'(c)=0 or does not exist, and c is in the domain of f, then c is a critical number. (Derivative is 0 or undefined)
Fundamental Theorem of Calculus #1
The definite integral of a rate of change is the total change in the original function.
-sin(x)
Combo Test for local extrema
If f'(c) = 0 and f"(c)<0, there is a local max on f at x=c. If f'(c) = 0 and f"(c)>0, there is a local min on f at x=c.
sec²(x)
L'Hopital's Rule
cos(x)
First Derivative Test for local extrema
sec(x)+C
-cot(x)+C
x+c
nx^(n-1)
Rolle's Theorem
Let f be continuous on [a,b] and differentiable on (a,b) and if f(a)=f(b) then there is at least one number c on (a,b) such that f'(c)=0 (If the slope of the secant is 0, the derivative must = 0 somewhere in the interval).
The position function OR s(t)
Area under a curve
ln(sinx)+C = -ln(cscx)+C
Squeeze Theorem
-ln(cosx)+C = ln(secx)+C
hint: tanu = sinu/cosu
Alternative Definition of a Derivative
f '(x) is the limit of the following difference quotient as x approaches c
Intermediate Value Theorem
If f is continuous on [a,b] and k is a number between f(a) and f(b), then there exists at least one number c such that f(c)=k
tan(x)+C
-cos(x)+C
Exponential growth (use N= )
dy/dx
Point of inflection at x=k
ln(cscx+cotx)+C = -ln(cscx-cotx)+C
0
f is continuous at x=c if...
cf'(x)
-csc²(x)
Formula for Washer Method
Axis of rotation is not a boundary of the region.
ln(secx+tanx)+C = -ln(secx-tanx)+C
If f and g are inverses of each other, g'(x)
sin(x)+C
1
Formula for Disk Method
Axis of rotation is a boundary of the region.
uvw'+uv'w+u'vw
Section 2
(50 cards)
cos(3π/2)
0
sin(5π/4)
−√2/2
cos(5π/3)
1/2
Cosine function
D: (-∞,+∞) R: [-1,1]
Constants in integrals
Sine function
D: (-∞,+∞) R: [-1,1]
Natural log function
D: (0,+∞) R: (-∞,+∞)
Derivative of eⁿ
cos(3π/4)
−√2/2
sin(7π/6)
−1/2
Opposite Antiderivatives
sin(5π/6)
1/2
Derivative of ln(u)
cos(7π/4)
√2/2
Absolute value function
D: (-∞,+∞) R: [0,+∞)
sin(4π/3)
−√3/2
cos(2π)
1
sin(π/3)
√3/2
Inverse Sine Antiderivative
sin(π)
0
Inverse Tangent Antiderivative
cos(7π/6)
−√3/2
sin(2π/3)
√3/2
sin(3π/2)
−1
sin(π/2)
1
cos(5π/4)
−√2/2
Squaring function
D: (-∞,+∞) R: (o,+∞)
Reciprocal function
D: (-∞,+∞) x can't be zero R: (-∞,+∞) y can't be zero
Inverse Secant Antiderivative
cos(π/2)
0
cos(11π/6)
√3/2
Antiderivative of xⁿ
ln(a)*aⁿ+C
cos(2π/3)
−1/2
Square root function
D: (0,+∞) R: (0,+∞)
sin(3π/4)
√2/2
Identity function
D: (-∞,+∞) R: (-∞,+∞)
cos(π/3)
1/2
Adding or subtracting antiderivatives
Antiderivative of f(x) from [a,b]
cos(5π/6)
−√3/2
cos(π)
−1
cos(π/4)
√2/2
Exponential function
D: (-∞,+∞) R: (0,+∞)
sin(π/4)
√2/2
Cubing function
D: (-∞,+∞) R: (-∞,+∞)
cos(π/6)
√3/2
cos(4π/3)
−1/2
sin(π/6)
1/2
Logistic function
D: (-∞,+∞) R: (0, 1)
Section 3
(50 cards)
d/dx[secx]=
secxtanx
f(x) = e^(x) +2
Asymptote: y=2 Domain: (-∞, ∞)
d/dx[tan⁻¹x]=
1/(1+x^2)
f(x)=-lnx
Asymptote: x=0 Domain: (0, ∞)
What does the graph y = sin(x) look like?
f(x)= -2+lnx
Asymptote: x=0 Domain: (0, ∞)
∫1/(a^2+x^2)dx=
(1/a)(tan⁻¹(x/a)+C
Trig Identity: sin(2x)=
2sinxcosx
Integration by Parts: Choice of u
I = Inverse Trig Function L = Natural log (lnx) A = Algebraic Expression (x, x², x³...) T = Trig function (sinx, cosx) E = e^x
f(x)=ln(x-2)
Asymptote: x=2 Domain: (2, ∞)
∫1/√(1-x^2)dx=
sin⁻¹x+C
What does the graph y = cos(x) look like?
sin(5π/3)
−√3/2
What does the graph y = cot(x) look like?
What does the graph y = tan(x) look like?
Trig Identity: sin²x=
½(1-cos(2x))
d/dx[tanx]=
sec²x
f(x) = e^(x-2)
Asymptote: y=0 Domain: (-∞, ∞)
Trig Identity: sec²x=
tan²x+1
d/dx[a^x]=
a^x*lna
f(x)=e^(x+2)
Asymptote: y=0 Domain: (-∞, ∞)
What does the graph y = sec(x) look like?
d/dx[sin⁻¹x]=
1/√(1-x^2)
∫secxdx=
ln|secx+tanx|+C
f(x)=ln(-x)
Asymptote: x=0 Domain: (-∞, 0)
d/dx[cos⁻¹x]=
-1/√(1-x^2)
∫1/(1+x^2)dx=
tan⁻¹x+C
∫e^xdx=
e^x+C
sin(11π/6)
−1/2
Trig Identity: cos(2x)=
1-2sin²x = 2cos²x-1
What does the graph y = sin(x) look like?
What does the graph y = tan(x) look like?
∫a^xdx=
(a^x)/lna+C
d/dx[e^x]=
e^x
sin(2π)
0
What does the graph y = cot(x) look like?
d/dx[cotx]=
-csc²x
What does the graph y = sec(x) look like?
∫tanxdx=
ln|secx|+C
∫1/xdx=
ln|x|+C
What does the graph y = cos(x) look like?
What does the graph y = csc(x) look like?
d/dx[a^g(x)]=
g'(x)a^g(x)lna
d/dx[e^g(x)]=
g'(x)e^g(x)
f(x)=ln(x+2)
Asymptote: x=-2 Domain: (-2, ∞)
Trig Identity: 1=
cos²x+sin²x
What does the graph y = csc(x) look like?
sin(7π/4)
−√2/2
d/dx[cscx]=
-cscxcotx
Trig Identity: cos²x=
½(1+cos(2x))
Section 4
(27 cards)