# Calculus Final Exam Review By: Bryant Nelson. Common Trigonometric Values x-value0π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/62π sin(x)0½1½0-½-½0 cos(x)1½0-½-½0½1.

## Presentation on theme: "Calculus Final Exam Review By: Bryant Nelson. Common Trigonometric Values x-value0π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/62π sin(x)0½1½0-½-½0 cos(x)1½0-½-½0½1."— Presentation transcript:

Calculus Final Exam Review By: Bryant Nelson

Common Trigonometric Values x-value0π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/62π sin(x)0½1½0-½-½0 cos(x)1½0-½-½0½1

Special Trigonometric Limits lim h  0 sin(h) h = ? 1 lim h  0 cos(h) - 1 h = ? 0

Differentiation Rules Definition of a Derivative: f’(x) = lim Δx  0 f(x+Δx) – f(x) Δx

Differentiation Rules cont. Product Rule: (f·g)’ = f’·g + f·g’ Quotient Rule: (f/g)’ = (f’·g - f·g’)/g 2 Natural Log Rule: d/dx(ln(u)) = (1/u) ·du/dx Exponential Rules: d/dx(℮ u ) = (℮ u ) ·du/dx d/dx(b u ) = (b u ) ·ln(b)·du/dx ·

Differentiation Rules cont. d/dx(sin(u)) =(cos(u))du/dx Trigonometric Rules d/dx(cos(u)) =(-sin(u))du/dx d/dx(tan(u)) =(sec 2 (u))du/dxd/dx(cot(u)) =(-csc 2 (u))du/dx d/dx(sec(u)) =(sec(u)tan(u))du/dxd/dx(csc(u)) =(-csc(u)cot(u))du/dx

Differentiation Rules cont. d/dx(sin -1 (u)) =(1/(√1-u 2 ))du/dx Inverse Trigonometric Rules d/dx(cos -1 (u)) =(-1/(√1-u 2 ))du/dx d/dx(tan -1 (u)) =(1/(1+u 2 ))du/dxd/dx(cot -1 (u)) =(-1/(1+u 2 ))du/dx d/dx(sec -1 (u)) =(1/(|u|·√u 2 -1))du/dxd/dx(csc -1 (u)) =(-1/(|u|·√u 2 -1))du/dx

Integration Rules Power Rules: ∫(u n ·du) = (u n+1 )/(n+1) +C; only while n ≠ -1 ∫(u -1 ·du) = ln(|u|) +C Exponential Rules: ∫(℮ u ·du) = ℮ u +C ∫(b u ·du) = b u /ln(b) - u +C, b≠1 Logarithmic Rule: ∫(ln(u)·du) = u·ln(u) - u +C, u>0

Integration Rules cont. ∫(sin(u)·du) =-cos(u) + C Trigonometric Rules ∫(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. ∫(sec 2 (u) ·du) =tan(u) + C∫(csc 2 (u) ·du) =-cot(u) + C ∫(sec(u)tan(u) ·du) =sec(u) + C∫(csc(u)cot(u) ·du) =-csc(u) + C

Summation Formulas n ∑ n K=1 1 = (n(n+1))/2 ∑ n K=1 k = (n(n+1)(2n+1))/6 ∑ n K=1 k 2 = (n 2 (n+1) 2 )/4 ∑ n K=1 k 3 =∑ n K=1 c·a k =∑ n K=1 akak c·c·

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