# Limits, Derivatives, Integrals (ch 1-5)

## 61 terms

log₆x=n = 6ⁿ=x

log₆x=n
6ⁿ=x
n⋅ln(6)=ln(x)
n= ln(x) ÷ ln(6)

### instantaneous velocity

lim Avg. Velocity
t→t₁

### Direct substitution property

If f is a polynomial or a rational function and a in the domain of f, then lim(x→a) f(x) = f(a)

### The Squeeze Theorem

lim(x→a) f(x) = L if lim(x→a₋) f(x)=l =lim(x→a+) f(x)
1) write inequality (-1<f(x)<1)
2) mult. by coeff. in f(x)
3) find limits of either side of f(x)
4) if they are equal then f(x) is the same as them by the squeeze theorem

### Precise Definition of a Limit

δ ∋ l f(x)-L l < ∈, ∀ x ∋ 0 < l x-al < δ
1) graph function with a and L and ∈s and x1 and x2
2) take lx₁-al and lx₂-al
3) take the bigger value

### Precise Definition of a Limit Proof

⌊f(x) -L⌋ < ∈ where ⌊x-a⌋ < δ
If answer has x in it still, restrict values around a

### L'hopital's rule

Use when both numerator and denominator equal 0 or ∞ when a is plugged in
1) take the derivative of both the numerator and the denominator
2) Repeat until you're able to plug in a

### Functions that are continous everywhere

1) polynomials
2) rational functions
3) root functions
4) trig functions

### Proofs by IVT

1) Is f(x) continuous in interval?
2) Plug in endpoints of interval
3) There exists c such that f(c) = 0

### Horizontal asymptotes

- if degree of numerator = degree of denominator, then coefficients of x with the highest degrees
- if degree of numerator > degree of denominator, then no horizontal asymptote
- if degree of numerator < degree of denominator, then horizontal asymptote = 1

### Tangent line to a curve

m = lim(x→a) [f(x)-f(a)]÷[x-a]
y-f(a)=f'(a)(x-a)

### Definition of Derivative

f'(a) = lim(h→0) [f(a+h)-f(a)]÷h if the limit exists

### Definition of Derivative as a function

f'(x) = lim(h→0) [f(x+h)-f(x)]÷h

### The Quotient Rule

f'[(f(x)⋅g(x)]=[g(x)⋅f'(x) - f(x)⋅g'(x)]÷[g(x)]²

cosx

-sinx

sec²x

-csc²x

secx⋅tanx

-csc²x

### Chain Rule

dy/dy f(g(x)) = f'(g(x))⋅g'(x)

### derivative of y=a^f(x)

y'=[a^f(x)]⋅ln(a)⋅f'(x)

### Implicit Differentiation

1) solve for y'
2) find the equation of tangent line (sometimes)

1/(√(1-x²)

-1/(√(1-x²)

1/(x⋅√(x²-1)

-1/(x⋅√(x²-1)

1/(1+x²)

-1/(1+x²)

1/xlna

### Logarithmic Differentiation

1) take ln of both sides
2) bring exponent down
3) take the derivative implicitly
4) solve for y'
5) substitute original function in for the ys (NOT the y's)

y(t)=Ce^kt

### Linear Approximation Equation

f(x)-f(a) = f'(a)(x-a)
f(x)=f'(x₁)(x-x₁)+f(x₁)

### Linear Approximation

1) Pick a function
2) Pick a number for x₁
3) find slope of tangent line; plug in x₁
4) Find coordinates (using f(x) and x₁)
5) Use linear approximation equation
f(x)-f(a) = f'(a)(x-a)

### Critical Number

a number c in the domain for f such that either f'(c)=0 or f'(c) does not exists (maximums and minimums are critical numbers)

### How to evaluate/graph a function for max, mins, etc

1) Take derivative
2) set derivative equal to zero to find critical numbers
3) Set up chart using endpoints and critical number, with values in between (no f(x) value needed for in-between values)
4)plug in to find f(x) values and f'(x) and f''(x) directions (positive or negative)
5) graph

### local max

f'(x) goes + to 0 to -

### local min

f'(x) goes - to 0 to +

### Rolle's Theorem

1) Use IVT to prove f(x) has at least one real root
2) continuous? differentiable? f(a)=f(b)?
3) f(x)=f'(x)?

### The Mean Value Theorem

1) Continuous and differentiable on closed interval?
2) f'(c) = [f(b)-f(a)]÷[b-a]
3) If x≠0, then yes only one root

ln(x)+c

### antiderivative a^x

[a^x]÷[ln(a)] + c

### First Derivative Test

if f'(x) changes from + to -, f has local max at c
if f'(x) changes from - to +, f has local min at c

### Concavity

+f"(x) = concave up
-f"(x) = concave down

### Second Derivative Test

If f"(s) is - at a critical point, its a local max
if f"(x) is + at a critical point, its a local min

### Optimization

1) pick an f(x) and an equation
i.e., f(x)=xy and x+y=20
2) combine
i.e., y-20-x, so f(x)=20x-x²
3) Find max using f'(x)
i.e., f'(x) =20-2x=0, x=10 (critical number(
4) analyze the function
max at 10

n

[n(n+1)]÷2

[n(n+1)(2n+1)]÷6

[[n(n+1)]÷2]²

### Riemann Sums

1) Sketch graph with rectangle
2) a=bh for each rectangle, where h=f(b) (for right-hand, its b+1)

### lim(n→∞) [16n³+24n²+8n]÷[6n³]

8/3 because n's are same coefficients: 16/6=8/3

∆x=(b-a)/n

xi=a+i∆x

### Fundamental Theorem of Calculus

if f is continuous on [a,b], then
1) if g(x) = ∫(a to x) f(t)dt, then g'(x)=f(x)
2) ∫(from a to b) f(x)dx = F(x)l(a to b) = f(b)-f(a)

∫xdx=(x²)/2 +c

### Total Change Theorem

The integral of a rate of change is called the total change: ∫(from a to b) F'(x)dx = F(b)-F(a)
-find anti-derivatives

[√(1-x³)] ⋅-3x²

### U-substitution indefinite integrals

1) set u = innermost function, du = u'⋅dx
2) Substitute du and u into integral
3) multiply the outside of the integral by fraction to balance du if necessary (if x, solve x for u first)
4) antiderivative
5) substitute back in

### U-subs with definite integrals

1) do u-subs normally (ignore bounds)
2) after taking antiderivative, plug bounds in using fundamental theorem of calculus part 2 (f(b)-f(a))