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Chapter 2: Limits and Derivatives
2.1 The Tangent and Velocity Problems
2.2 The Limit of a Function
Limits: lim f(x) = L
x→a
One-sided Limits: lim f(x) = L
x→a−
x=aisavertical asymptote of y = f(x) if at least one one-sided limit as x approaches a is ±∞.
2.3 Calculating Limits Using the Limit Laws
If lim f(x) and lim g(x) exist (particularly not ±∞):
x→a x→a
lim[f(x)+g(x)] = lim f(x)+ lim g(x)
x→a x→a x→a
lim[f(x)−g(x)] = lim f(x)− lim g(x)
x→a x→a x→a
lim[cf(x)] = c lim f(x) where c is a constant
x→a x→a
lim[f(x)g(x)] = lim f(x)· lim g(x)
x→a x→a x→a
f(x) lim f(x)
lim[ ] = x→a if limx→ag(x) 6= 0
x→a g(x) lim g(x)
h x→a i
n n
lim[f(x)] = lim f(x) where n is a positive integer
x→a x→a
p q
lim n f(x) = n lim f(x), n is a pos. int. (if n is even, assume lim f(x) > 0)
x→a x→a x→a
Squeeze Theorem: If f(x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a and lim f(x) =
lim h(x) = L, then lim g(x) = L. x→a
x→a x→a
2.5 Continuity
Afunction f(x) is continuous at a if lim f(x) = f(a).
x→a
Polynomials, exponentials, logarithms, roots, trig functions, inverse trig functions and rational func-
tions are all continuous at each point in their domains.
If f, g are continuous at a, c constant, then f +g, f −g, fg, cf, fg, f if g(a) 6= 0 are continuous at a.
g
If g is continuous at a and f is continuous at g(a), then f ◦ g is continuous at a.
Intermediate Value Theorem: Suppose that f is continuous on [a,b] and let N be any number between
f(a) and f(b), where f(a) 6= f(b). Then there exists a number c in (a,b) such that f(c) = N.
2.6 Limits at Infinity: Horizontal Asymptotes
y = L is a horizontal asymptote of the curve y = f(x) if lim = L or lim =L.
x→∞ x→−∞
2.7 Derivatives and Rates of Change
The tangent line to y = f(x) at (a,f(a)) is the line through (a,f(a)) whose slope is equal to f′(a), the
derivative of f at a.
Point-slope formula: The equation of a line with slope f′(a) at the point (a,f(a)) is y − f(a) =
f′(a)(x−a)
The following terms mean the same thing: the derivative, the slope of the line tangent to the curve,
and the instantaneous rate of change.
2.8 The Derivative as a Function
Definition of the derivative: f′(x) = lim f(x+h)−f(x) (provided the limit exists)
h→0 h
Differentiation is the process of taking a derivative, and a function f is differentiable at a if f′(a)
exists.
If f is differentiable at a then f is continuous at a.
Given y = f(x), we can denote its derivative as f′(x), y′, d f(x), or d y = dy (Leibniz notation).
dx dx dx
Chapter 3: Differentiation Rules
3.1 Derivatives of Polynomials and Exponential Functions
d (c) = 0 (where c is any constant)
dx
d (xn) = nxn−1 (where n is any constant)
dx
d x x d x x
dx(e ) = e dx(a ) = a lna
If f and g are both differentiable and c is constant, then
d [cf(x)] = cf′(x) d [f(x)±g(x)] = f′(x)±g′(x)
dx dx
3.2 The Product and Quotient Rules
If f and g are both differentiable, then
d [f(x)g(x)] = f(x)g′(x)+g(x)f′(x) d f(x) = g(x)f′(x)−f(x)g′(x)
dx dx g(x) (g(x))2
3.3 Derivatives of Trigonometric Functions
d d d 2
dx(sinx) = cosx dx(cosx) = −sinx dx(tanx) = sec x
d (cscx) = −csccotx d (secx) = sectanx d (cotx) = −csc2x
dx dx dx
3.4 The Chain Rule
If g is differentiable at x and f is differentiable at g(x), then the composite F = f ◦ g defined by
F(x) = f(g(x)) is differentiable at x, and F′ is given by F′(x) = f′(g(x)) · g′(x).
Also written as d f(g(x)) = f′(g(x))·g′(x) or dy = dy · du
dx dx du dx
3.5 Implicit Differentiation
Implicit Differentiation: Differentiate both sides of the equation y = f(x) with respect to x and then
solve the resulting equation for y′ = dy.
dx
d (arcsinx) = √ 1 d (arccosx) = −√ 1 d (arctanx) = 1
dx 1−x2 dx 1−x2 dx 1+x2
d (arccscx) = − √ 1 d (arcsecx) = √ 1 d (arccotx) = − 1
dx 2 dx 2 dx 1+x2
x x −1 x x −1
3.6 Derivatives of Logarithmic Functions
d (lnx) = 1 d (log x) = 1 d (ln|x|) = 1
dx x dx b xlnb dx x
Logarithmic Differentiation: Take natural logarithms of both sides of an equation y = f(x) and use
the logarithm laws to simplify, and then differentiate implicitly with respect to x to solve for y′.
3.7 Rates of Change in the Natural and Social Sciences
′
If s(t) is the position function of a particle at time t, then v = s (t) represents the instantaneous
velocity, and a = v′(t) = s′′(t) represents the acceleration.
3.8 Exponential Growth and Decay
dy kx
The differential equation dx = ky only has solutions of the form Ce , where C is a constant. More
kx
precisely, the solutions are y(x) = y(0)e .
Herewesaythat“thegrowth/decayrateisproportionaltothesize/mass”or“therelativegrowth/decay
rate is constant.”
The half-life is the time required for half of a quantity to decay.
3.9 Related Rates
Formulas to know:
– Pythagorean theorem a2 +b2 = c2
′
– Similar triangles a = a
b b′
2 2 2 2 2 2
– Trig identities: sin x+cos x = 1, tan x+1 = sec x, 1+cot x = csc x, sin(2x) = 2sinxcosx,
2 2
cos(2x) = cos x−sin x
– Evaluation of trigonometric functions at special angles (π, π, 0, etc.)
2
– Relationship of each of the six trigonometric functions to the hypotenuse and the opposite and
adjacent sides of a right triangle (SOH-CAH-TOA)
– Circumference (2πr) and diameter (2π) of a circle
– Areas of rectangle (A = l · w), circle (A = πr2), triangle (A = 1b · h)
2
– Volumes of box (V = l ·w ·h), sphere (V = 4πr2), cone (V = 1πr2h),
3 p 3
– Distance between points (x ,y ) and (x ,y ) is D = (x −x )2+(y −y )2
1 1 2 2 2 1 2 1
Problem-solving strategies:
1. Read the problem and draw a diagram, introducing notation for quantities.
2. Express the given information and the required rate in terms of derivatives, and write an equation
that relates the various quantities of the problem (try to eliminate variables by substitution using
geometry).
3. Use the chain rule to differentiate both sides of the equation with respect to t.
4. Substitute the given information into the resulting equation and solve for the unknown rate.
3.10 Linear Approximations and Differentials
Linear approximation/Tangent line approximation: f(x) ≈ L(x) = f(a)+f′(a)(x−a)
3.11 Hyperbolic Functions
x −x x −x
sinhx = e −e coshx = e +e
2 2
tanhx = sinhx cschx = 1 sechx = 1 cothx = coshx
coshx sinhx coshx sinhx
d (sinhx) = coshx d (coshx) = sinhx d (tanhx) = sech2x
dx dx dx
d (cschx) = −cschxcothx d (sechx) = −sechxtanhx d (cothx) = −csch2x
dx dx dx
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