Section 1
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Formal definition of derivative
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Mar 1, 2020
Cards (75)
Section 1
(50 cards)
Formal definition of derivative
Linearization
use tangent line to approximate values of the function
y = ln(x)/x², state rule used to find derivative
quotient rule
y = x cos(x), state rule used to find derivative
product rule
Quotient Rule
(uv'-vu')/v²
y = csc(x), y' =
y' = -csc(x)cot(x)
area above x-axis is
positive
Chain Rule
f '(g(x)) g'(x)
If g(x) = ∫ f(t) dt on interval 2 to x, then g'(x) =
g'(x) = f(x)
absolute value of velocity
speed
When f '(x) is positive, f(x) is
increasing
Product Rule
uv' + vu'
Instantaneous Rate of Change (words)
Slope of tangent line at a point, value of derivative at a point
If f '(x) = 0 and f"(x) < 0,
f(x) has a relative maximum
To find absolute maximum on closed interval [a, b], you must consider...
critical points and endpoints
Average Rate of Change (words)
Slope of secant line between two points, use to estimate instantanous rate of change at a point.
y = cot⁻¹(x), y' =
y' = -1/(1 + x²)
y = tan⁻¹(x), y' =
y' = 1/(1 + x²)
When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x) has a
point of inflection
right riemann sum
use rectangles with right-endpoints to evaluate integrals (estimate area)
When f '(x) changes from positive to negative, f(x) has a
relative maximum
When f '(x) changes from negative to positive, f(x) has a
relative minimum
y = cos²(3x)
chain rule
If f '(x) = 0 and f"(x) > 0,
f(x) has a relative minimum
y = sec(x), y' =
y' = sec(x)tan(x)
Particle is moving to the right/up
velocity is positive
slope of horizontal line
zero
mean value theorem
if f(x) is continuous and differentiable, slope of tangent line equals slope of secant line at least once in the interval (a, b) f '(c) = [f(b) - f(a)]/(b - a)
When f '(x) is increasing, f(x) is
concave up
left riemann sum
use rectangles with left-endpoints to evaluate integral (estimate area)
[(h1 - h2)/2]*base
area of trapezoid
Fundamental Theorem of Calculus
∫ f(x) dx on interval a to b = F(b) - F(a)
application of Intermediate Value Theorem
If f(1)=-4 and f(6)=9, then there must be a x-value between 1 and 6 where f crosses the x-axis.
When f '(x) is decreasing, f(x) is
concave down
indefinite integral
no limits, find antiderivative + C, use inital value to find C
Particle is moving to the left/down
velocity is negative
y = cot(x), y' =
y' = -csc²(x)
y = e^x, y' =
y' = e^x
y = cos⁻¹(x), y' =
y' = -1/√(1 - x²)
When is a function not differentiable
corner, cusp, vertical tangent, discontinuity
y = a^x, y' =
y' = a^x ln(a)
y = cos(x), y' =
y' = -sin(x)
trapezoidal rule
use trapezoids to evaluate integrals (estimate area)
y = sin(x), y' =
y' = cos(x)
y = sin⁻¹(x), y' =
y' = 1/√(1 - x²)
area below x-axis is
negative
When f '(x) is negative, f(x) is
decreasing
area under a curve
∫ f(x) dx integrate over interval a to b
y = tan(x), y' =
y' = sec²(x)
definite integral
has limits a & b, find antiderivative, F(b) - F(a)
Section 2
(25 cards)