Transcript Calculus Final Exam Review

Calculus Final Exam Review
By: Bryant Nelson
Common Trigonometric Values
x-value
0 π/6
sin(x)
0 ½
cos(x)
1
π/3
π/2
2π/3
1
½
0
-½
5π/6
π
7π/6
½
0
-½
-1
4π/3
3π/2
5π/3
-1
-½
0
½
11π/6
2π
-½
0
1
Special Trigonometric Limits
lim
sin(h)
h0
=?
1
=?
0
h
lim
cos(h) - 1
h0
h
Differentiation Rules
Definition of a Derivative:
f’(x)
=
lim
f(x+Δx) – f(x)
Δx0
Δx
Differentiation Rules cont.
Product Rule:
(f·g)’ = f’·g + f·g’
Quotient Rule: (f/g)’ = (f’·g - f·g’)/g2
Natural Log Rule:
Exponential Rules:
d/dx(ln(u)) = (1/u)·du/dx
d/dx(℮u) = (℮u·)·du/dx
d/dx(bu) = (bu) ·ln(b)·du/dx
Differentiation Rules cont.
Trigonometric Rules
d/dx(sin(u)) =
(cos(u))du/dx
d/dx(cos(u)) =
(-sin(u))du/dx
d/dx(tan(u)) =
(sec2(u))du/dx
d/dx(cot(u)) =
(-csc2(u))du/dx
d/dx(sec(u)) =
(sec(u)tan(u))du/dx
d/dx(csc(u)) = (-csc(u)cot(u))du/dx
Differentiation Rules cont.
Inverse Trigonometric Rules
d/dx(sin-1(u)) = (1/(√1-u2))du/dx
d/dx(cos-1(u)) = (-1/(√1-u2))du/dx
d/dx(tan-1(u)) = (1/(1+u2))du/dx
d/dx(cot-1(u)) = (-1/(1+u2))du/dx
d/dx(sec-1(u)) = (1/(|u|·√u2-1))du/dx
d/dx(csc -1(u)) =(-1/(|u|·√u2-1))du/dx
Integration Rules
Power Rules:
∫(un·du) = (un+1)/(n+1) +C; only while n ≠ -1
∫(u-1·du) = ln(|u|) +C
Exponential Rules:
∫(℮u·du) = ℮u +C
Logarithmic Rule:
∫(bu·du) = bu/ln(b) - u +C,
b≠1
∫(ln(u)·du) = u·ln(u) - u +C, u>0
Integration Rules cont.
Trigonometric Rules
∫(sin(u)·du) =
-cos(u) + C
∫(cos(u) ·du) =
sin(u) + C
∫(tan(u) ·du) =
-ln(|cos(u)|) + C
∫(cot(u) ·du) =
ln(|sin(u)|) + C
∫(sec(u) ·du) =
ln(|sec(u) + tan(u)|) + C
∫(csc(u) ·du) =ln(|csc(u) + cot(u)|) + C
Integration Rules cont.
Trigonometric Rules cont.
∫(sec2(u) ·du) = tan(u) + C
∫(csc2(u) ·du) = -cot(u) + C
∫(sec(u)tan(u) ·du) = sec(u) + C
∫(csc(u)cot(u) ·du) = -csc(u) + C
Summation Formulas
n
∑1=
n
K=1
n
∑k=
(n(n+1))/2
K=1
n
∑ k2 = (n(n+1)(2n+1))/6
K=1
n
∑ k3 = (n2(n+1)2)/4
K=1
n
n
∑ c·ak = c·∑ ak
K=1
K=1