Section 1
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-csc(x)+C
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Cards (176)
Section 1
(50 cards)
-csc(x)+C
cf'(x)
Point of inflection at x=k
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.
uvw'+uv'w+u'vw
1
-csc²(x)
Global Definition of a Derivative
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)
Horizontal Asymptote
ln(x)+C
nx^(n-1)
f is continuous at x=c if...
1
Alternative Definition of a Derivative
f '(x) is the limit of the following difference quotient as x approaches c
sec(x)tan(x)
f'(x)-g'(x)
Mean Value Theorem for integrals or the average value of a functions
The position function OR s(t)
sec²(x)
Fundamental Theorem of Calculus #2
Squeeze Theorem
First Derivative Test for local extrema
-ln(cosx)+C = ln(secx)+C
hint: tanu = sinu/cosu
Exponential growth (use N= )
ln(sinx)+C = -ln(cscx)+C
x+c
f'(x)+g'(x)
dy/dx
-cot(x)+C
Formula for Disk Method
Axis of rotation is a boundary of the region.
ln(secx+tanx)+C = -ln(secx-tanx)+C
Area under a curve
Formula for Washer Method
Axis of rotation is not a boundary of the region.
L'Hopital's Rule
-sin(x)
Fundamental Theorem of Calculus #1
The definite integral of a rate of change is the total change in the original function.
0
tan(x)+C
sin(x)+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
cos(x)
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).
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.
sec(x)+C
If f and g are inverses of each other, g'(x)
f'(g(x))g'(x)
-cos(x)+C
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.
ln(cscx+cotx)+C = -ln(cscx-cotx)+C
Section 2
(50 cards)
cos(11π/6)
√3/2
Reciprocal function
D: (-∞,+∞) x can't be zero R: (-∞,+∞) y can't be zero
Absolute value function
D: (-∞,+∞) R: [0,+∞)
sin(π/3)
√3/2
Squaring function
D: (-∞,+∞) R: (o,+∞)
Inverse Tangent Antiderivative
cos(3π/2)
0
cos(5π/4)
−√2/2
Identity function
D: (-∞,+∞) R: (-∞,+∞)
Adding or subtracting antiderivatives
Cosine function
D: (-∞,+∞) R: [-1,1]
Inverse Secant Antiderivative
cos(5π/6)
−√3/2
sin(π/4)
√2/2
Antiderivative of xⁿ
Square root function
D: (0,+∞) R: (0,+∞)
sin(4π/3)
−√3/2
cos(5π/3)
1/2
cos(4π/3)
−1/2
sin(5π/4)
−√2/2
Cubing function
D: (-∞,+∞) R: (-∞,+∞)
sin(3π/4)
√2/2
sin(2π/3)
√3/2
cos(2π/3)
−1/2
Antiderivative of f(x) from [a,b]
Logistic function
D: (-∞,+∞) R: (0, 1)
cos(π)
−1
ln(a)*aⁿ+C
cos(3π/4)
−√2/2
cos(π/4)
√2/2
Exponential function
D: (-∞,+∞) R: (0,+∞)
Natural log function
D: (0,+∞) R: (-∞,+∞)
cos(7π/4)
√2/2
sin(π)
0
sin(7π/6)
−1/2
sin(3π/2)
−1
sin(π/6)
1/2
cos(π/3)
1/2
Inverse Sine Antiderivative
Derivative of eⁿ
cos(π/6)
√3/2
cos(2π)
1
Derivative of ln(u)
cos(π/2)
0
Sine function
D: (-∞,+∞) R: [-1,1]
Opposite Antiderivatives
sin(5π/6)
1/2
cos(7π/6)
−√3/2
sin(π/2)
1
Constants in integrals
Section 3
(50 cards)
f(x)=e^(x+2)
Asymptote: y=0 Domain: (-∞, ∞)
What does the graph y = tan(x) look like?
What does the graph y = sin(x) look like?
What does the graph y = csc(x) look like?
f(x) = e^(x-2)
Asymptote: y=0 Domain: (-∞, ∞)
What does the graph y = sin(x) look like?
d/dx[tanx]=
sec²x
d/dx[cscx]=
-cscxcotx
∫1/(a^2+x^2)dx=
(1/a)(tan⁻¹(x/a)+C
sin(7π/4)
−√2/2
Trig Identity: cos²x=
½(1+cos(2x))
d/dx[secx]=
secxtanx
d/dx[a^x]=
a^x*lna
What does the graph y = sec(x) look like?
sin(5π/3)
−√3/2
∫e^xdx=
e^x+C
∫1/xdx=
ln|x|+C
Trig Identity: 1=
cos²x+sin²x
sin(11π/6)
−1/2
d/dx[cos⁻¹x]=
-1/√(1-x^2)
f(x)=ln(x-2)
Asymptote: x=2 Domain: (2, ∞)
What does the graph y = sec(x) look like?
∫secxdx=
ln|secx+tanx|+C
sin(2π)
0
f(x)=ln(-x)
Asymptote: x=0 Domain: (-∞, 0)
∫1/√(1-x^2)dx=
sin⁻¹x+C
f(x)= -2+lnx
Asymptote: x=0 Domain: (0, ∞)
d/dx[e^g(x)]=
g'(x)e^g(x)
d/dx[a^g(x)]=
g'(x)a^g(x)lna
What does the graph y = cot(x) look like?
d/dx[cotx]=
-csc²x
∫tanxdx=
ln|secx|+C
Trig Identity: sin²x=
½(1-cos(2x))
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
Trig Identity: cos(2x)=
1-2sin²x = 2cos²x-1
d/dx[e^x]=
e^x
What does the graph y = tan(x) look like?
What does the graph y = csc(x) look like?
What does the graph y = cot(x) look like?
∫a^xdx=
(a^x)/lna+C
∫1/(1+x^2)dx=
tan⁻¹x+C
d/dx[sin⁻¹x]=
1/√(1-x^2)
f(x)=-lnx
Asymptote: x=0 Domain: (0, ∞)
Trig Identity: sec²x=
tan²x+1
f(x) = e^(x) +2
Asymptote: y=2 Domain: (-∞, ∞)
What does the graph y = cos(x) look like?
d/dx[tan⁻¹x]=
1/(1+x^2)
What does the graph y = cos(x) look like?
f(x)=ln(x+2)
Asymptote: x=-2 Domain: (-2, ∞)
Trig Identity: sin(2x)=
2sinxcosx
Section 4
(26 cards)