To differentiate logarithmic functions with bases other than e, use
|
|
|
- Eleanore Lyons
- 8 years ago
- Views:
Transcription
1 To ifferentiate logarithmic functions with bases other than e, use 1
2 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b
3 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b Example Fin the erivative of y = log 2 (5x 3 ).
4 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b Example Fin the erivative of y = log 2 (5x 3 ). Solution y x = x ( ) ln 5x 3 ln 2
5 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b Example Fin the erivative of y = log 2 (5x 3 ). Solution y x = x ( ) ln 5x 3 ln 2 = x ( 1 ln 2 ln 5x3 )
6 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b Example Fin the erivative of y = log 2 (5x 3 ). Solution y x = x ( ) ln 5x 3 ln 2 = x ( 1 ln 2 ln 5x3 ) = 1 ln 2 (ln ln x) x properties of log
7 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b Example Fin the erivative of y = log 2 (5x 3 ). Solution y x = x ( ) ln 5x 3 ln 2 = x ( 1 ln 2 ln 5x3 ) = = 1 ln 2 1 ln 2 (ln ln x) x ( ) x properties of log
8 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b Example Fin the erivative of y = log 2 (5x 3 ). Solution y x = x ( ) ln 5x 3 ln 2 = x ( 1 ln 2 ln 5x3 ) = = 1 ln 2 1 ln 2 (ln ln x) x ( ) x properties of log = 3 x ln 2
9 2 Exponential functions of other bases To ifferentiate f (x) = b x where b e
10 2 Exponential functions of other bases To ifferentiate f (x) = b x where b e Metho 1 Express b x using exponential with base e. Metho 2 Use a technique calle logarithmic ifferentiation.
11 2 Exponential functions of other bases To ifferentiate f (x) = b x where b e Metho 1 Express b x using exponential with base e. y = b x Metho 2 Use a technique calle logarithmic ifferentiation.
12 2 Exponential functions of other bases To ifferentiate f (x) = b x where b e Metho 1 Express b x using exponential with base e. y = b x ln y = ln b x = x ln b Metho 2 Use a technique calle logarithmic ifferentiation.
13 2 Exponential functions of other bases To ifferentiate f (x) = b x where b e Metho 1 Express b x using exponential with base e. y = b x ln y = ln b x = x ln b (ln b)x y = e Metho 2 Use a technique calle logarithmic ifferentiation.
14 2 Exponential functions of other bases To ifferentiate f (x) = b x where b e Metho 1 Express b x using exponential with base e. y = b x ln y = ln b x = x ln b (ln b)x y = e Metho 2 Use a technique calle logarithmic ifferentiation. Both methos nee chain rule.
15 3 Chapter 9: More Differentiation Chain Rule Implicit Differentiation More Curve Sketching More Extremum Problems Objectives To use Chain Rule to o ifferentiation. To use Implicit Differentiation to fin y x. To apply ifferentiation.
16 4 Up to this moment, can ifferentiate simple functions like (1) f (x) = x (2) f (x) = x 1 x + 1 (3) f (x) = sin x (4) f (x) = e x + 2 tan x (5) f (x) = ln x cos x ex x using simple rules an formulas erive in the last few chapters.
17 5 How about (1) g(x) = sin(x 2 )? (2) g(x) = e x2 +1? (3) g(x) = ln(1 + 2x)?
18 5 How about (1) g(x) = sin(x 2 )? (2) g(x) = e x2 +1? (3) g(x) = ln(1 + 2x)? Nee the chain rule most important rule for fining erivatives, use for ifferentiating composite functions.
19 5 How about (1) g(x) = sin(x 2 )? g(x) = sin f (x) where f (x) = x 2 (2) g(x) = e x2 +1? (3) g(x) = ln(1 + 2x)? Nee the chain rule most important rule for fining erivatives, use for ifferentiating composite functions.
20 5 How about (1) g(x) = sin(x 2 )? g(x) = sin f (x) where f (x) = x 2 (2) g(x) = e x2 +1? g(x) = e f (x) where f (x) = x (3) g(x) = ln(1 + 2x)? Nee the chain rule most important rule for fining erivatives, use for ifferentiating composite functions.
21 5 How about (1) g(x) = sin(x 2 )? g(x) = sin f (x) where f (x) = x 2 (2) g(x) = e x2 +1? g(x) = e f (x) where f (x) = x (3) g(x) = ln(1 + 2x)? g(x) = ln f (x) where f (x) = 1 + 2x Nee the chain rule most important rule for fining erivatives, use for ifferentiating composite functions.
22 Composition of Functions Recall (g f )(x) = g ( f (x) ) 6
23 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) (2) (g f )(x)
24 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) Solution ( f g)(x) = f ( g(x) ) (2) (g f )(x)
25 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) Solution ( f g)(x) = f ( g(x) ) = f (x 2 ) (2) (g f )(x)
26 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) Solution ( f g)(x) = f ( g(x) ) = f (x 2 ) = sin(x 2 ) = sin x 2 (2) (g f )(x)
27 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) Solution ( f g)(x) = f ( g(x) ) = f (x 2 ) = sin(x 2 ) = sin x 2 (2) (g f )(x) Solution (g f )(x) = g ( f (x) )
28 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) Solution ( f g)(x) = f ( g(x) ) = f (x 2 ) = sin(x 2 ) = sin x 2 (2) (g f )(x) Solution (g f )(x) = g ( f (x) ) = g(sin x)
29 6 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Let f (x) = sin x an g(x) = x 2. Fin (1) ( f g)(x) Solution ( f g)(x) = f ( g(x) ) = f (x 2 ) = sin(x 2 ) = sin x 2 (2) (g f )(x) Solution (g f )(x) = g ( f (x) ) = g(sin x) = (sin x) 2 = sin 2 x
30 7 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Express e x2 +1 as composition of two functions.
31 7 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Express e x2 +1 as composition of two functions. Solution Let f (x) = x an
32 7 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Express e x2 +1 as composition of two functions. Solution Let f (x) = x an e x2 +1 = e f (x)
33 7 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Express e x2 +1 as composition of two functions. Solution Let f (x) = x an g(x) = e x. Then e x2 +1 = e f (x)
34 7 Composition of Functions Recall (g f )(x) = g ( f (x) ) Example Express e x2 +1 as composition of two functions. Solution Let f (x) = x an g(x) = e x. Then e x2 +1 = e f (x) = g ( f (x) )
35 8 Chain Rule If y is a ifferentiable function of u an u is a ifferentiable function of x, then y is a ifferentiable function of x an
36 8 Chain Rule If y is a ifferentiable function of u an u is a ifferentiable function of x, then y is a ifferentiable function of x an y x = y u u x
37 8 Chain Rule If y is a ifferentiable function of u an u is a ifferentiable function of x, then y is a ifferentiable function of x an y x = y u u x Iea of proof y x = lim x 0 y x lim h 0 f (x + h) f (x) h
38 8 Chain Rule If y is a ifferentiable function of u an u is a ifferentiable function of x, then y is a ifferentiable function of x an y x = y u u x Iea of proof y x = lim x 0 y x lim h 0 f (x + h) f (x) h ( y = lim x 0 u u ) x
39 8 Chain Rule If y is a ifferentiable function of u an u is a ifferentiable function of x, then y is a ifferentiable function of x an y x = y u u x Iea of proof y x = lim x 0 y x lim h 0 f (x + h) f (x) h ( y = lim x 0 u u ) x = lim u 0 y u lim x 0 u x
40 8 Chain Rule If y is a ifferentiable function of u an u is a ifferentiable function of x, then y is a ifferentiable function of x an y x = y u u x Iea of proof y x = lim x 0 y x lim h 0 f (x + h) f (x) h ( y = lim x 0 u u ) x = lim u 0 = y u u x y u lim x 0 u x
41 9 Chain Rule in alternative form Put y = f (u) an u = g(x).
42 9 Chain Rule in alternative form Put y = f (u) an u = g(x). Then y = ( f g)(x) (composition of functions)
43 9 Chain Rule in alternative form Put y = f (u) an u = g(x). Then y = ( f g)(x) (composition of functions) ( f g) (x) = y x
44 9 Chain Rule in alternative form Put y = f (u) an u = g(x). Then y = ( f g)(x) (composition of functions) ( f g) (x) = y x = y u u x
45 9 Chain Rule in alternative form Put y = f (u) an u = g(x). Then y = ( f g)(x) (composition of functions) ( f g) (x) = y x = y u u x = f (u) g (x)
46 9 Chain Rule in alternative form Put y = f (u) an u = g(x). Then y = ( f g)(x) (composition of functions) ( f g) (x) = y x = y u u x = f (u) g (x) ( f g) (x) = f ( g(x) ) g (x)
47 10 Example Fin x (x2 + 5) 3 (1) without using chain rule; (2) using chain rule.
48 10 Example Fin x (x2 + 5) 3 (1) without using chain rule; (2) using chain rule. Solution (1) (without chain rule) Expaning (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) = x x x
49 10 Example Fin x (x2 + 5) 3 (1) without using chain rule; (2) using chain rule. Solution (1) (without chain rule) Expaning (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) Differentiating term by term: = x x x x (x2 + 5) 3 = x (x6 + 15x x ) = 6x x x = 6x x x
50 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule)
51 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x 2 + 5
52 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3.
53 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3.
54 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3. x (x2 + 5) 3 = y x
55 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3. x (x2 + 5) 3 = y x = y u u x chain rule
56 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3. x (x2 + 5) 3 = y x = y u u x = u u3 x (x2 + 5) chain rule
57 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3. x (x2 + 5) 3 = y x = y u u x = u u3 = 3u 2 2x x (x2 + 5) chain rule
58 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3. x (x2 + 5) 3 = y x = y u u x = u u3 = 3u 2 2x x (x2 + 5) = 3(x 2 + 5) 2 (2x) chain rule
59 11 Example Fin x (x2 + 5) 3 (2) using chain rule. Solution (2) (using chain rule) Put u = x an y = u 3. Then y = (x 2 + 5) 3. x (x2 + 5) 3 = y x = y u u x = u u3 = 3u 2 2x x (x2 + 5) = 3(x 2 + 5) 2 (2x) = 6x(x 2 + 5) 2 chain rule
60 Metho 1 Answer is 6x x x Metho 2 Answer is 6x(x 2 + 5) 2 12 Remark 1 The above two results are the same.
61 Metho 1 Answer is 6x x x Metho 2 Answer is 6x(x 2 + 5) 2 12 Remark 1 The above two results are the same. Remark 2 If change the function to y = (x 2 + 5) 1 3, first metho can t be applie.
62 Metho 1 Answer is 6x x x Metho 2 Answer is 6x(x 2 + 5) 2 12 Remark 1 The above two results are the same. Remark 2 If change the function to y = (x 2 + 5) 1 3, first metho can t be applie. Metho 1 (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) + 5 3
63 Metho 1 Answer is 6x x x Metho 2 Answer is 6x(x 2 + 5) 2 12 Remark 1 The above two results are the same. Remark 2 If change the function to y = (x 2 + 5) 3, 1 first metho can t be applie. Metho 1 (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) (x 2 + 5) 1 3 no way to expan
64 Metho 1 Answer is 6x x x Metho 2 Answer is 6x(x 2 + 5) 2 12 Remark 1 The above two results are the same. Remark 2 If change the function to y = (x 2 + 5) 3, 1 first metho can t be applie. Metho 1 (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) (x 2 + 5) 1 3 no way to expan Metho 2 Put u = x an y = u 3. Then y = (x 2 + 5) 3.
65 Metho 1 Answer is 6x x x Metho 2 Answer is 6x(x 2 + 5) 2 12 Remark 1 The above two results are the same. Remark 2 If change the function to y = (x 2 + 5) 3, 1 first metho can t be applie. Metho 1 (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) (x 2 + 5) 1 3 no way to expan Metho 2 Put u = x an y = u 3. Then y = (x 2 + 5) 3. Put u = x an y = u 1 3. Then y = (x 2 + 5) 3. 1
66 Metho 1 Answer is 6x x x 12 Metho 2 Answer is 6x(x 2 + 5) 2 Remark 1 The above two results are the same. Remark 2 If change the function to y = (x 2 + 5) 3, 1 first metho can t be applie. Metho 1 (x 2 + 5) 3 = (x 2 ) 3 + 3(x 2 ) 2 (5) + 3(x 2 )(5 2 ) (x 2 + 5) 1 3 no way to expan Metho 2 Put u = x an y = u 3. Then y = (x 2 + 5) 3. Put u = x an y = u 1 3. Then y = (x 2 + 5) 3. 1 Secon metho makes use of the chain rule together with the power rule.
67 13 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x x ln x = 1 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x x e f (x) = e f (x) ln[ f (x)] = 1 f (x) x f (x) x f (x) x f (x)
68 14 (3) x cos[ f (x)] = sin[ f (x)] x f (x) Proof
69 14 (3) x cos[ f (x)] = sin[ f (x)] x f (x) Proof Put u = f (x) an y = cos u.
70 14 (3) Proof x cos[ f (x)] = sin[ f (x)] x f (x) Put u = f (x) an y = cos u. Then y = cos f (x) an so
71 14 (3) Proof x cos[ f (x)] = sin[ f (x)] x f (x) Put u = f (x) an y = cos u. Then y = cos f (x) an so y cos[ f (x)] = x x
72 14 (3) Proof x cos[ f (x)] = sin[ f (x)] x f (x) Put u = f (x) an y = cos u. Then y = cos f (x) an so y cos[ f (x)] = x x = y u u x chain rule
73 14 (3) Proof x cos[ f (x)] = sin[ f (x)] x f (x) Put u = f (x) an y = cos u. Then y = cos f (x) an so y cos[ f (x)] = x x = y u u x = u cos u u x chain rule
74 14 (3) Proof x cos[ f (x)] = sin[ f (x)] x f (x) Put u = f (x) an y = cos u. Then y = cos f (x) an so y cos[ f (x)] = x x = y u u x = u cos u u x = sin u u x chain rule
75 14 (3) Proof x cos[ f (x)] = sin[ f (x)] x f (x) Put u = f (x) an y = cos u. Then y = cos f (x) an so y cos[ f (x)] = x x = y u u x = u cos u u x = sin u u x = sin[ f (x)] chain rule x f (x)
76 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof
77 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u.
78 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u. Then y = ln f (x) an so
79 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u. Then y = ln f (x) an so y ln[ f (x)] = x x
80 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u. Then y = ln f (x) an so y ln[ f (x)] = x x = y u u x chain rule
81 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u. Then y = ln f (x) an so y ln[ f (x)] = x x = y u u x = u ln u u x chain rule
82 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u. Then y = ln f (x) an so y ln[ f (x)] = x x = y u u x = u = 1 u u x ln u u x chain rule
83 15 (6) x ln[ f (x)] = 1 f (x) x f (x) Proof Put u = f (x) an y = ln u. Then y = ln f (x) an so y ln[ f (x)] = x x = y u u x = u = 1 u u x = 1 f (x) ln u u x x f (x) chain rule
84 16 Simple Form General Form x xr = rx r 1
85 16 Simple Form x xr = rx r 1 General Form x [ f (x)]r = r[ f (x)] r 1 x f (x)
86 16 Simple Form x xr = rx r 1 x sin x = cos x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x)
87 16 Simple Form x xr = rx r 1 x sin x = cos x General Form x [ f (x)]r = r[ f (x)] r 1 x sin[ f (x)] = x f (x)
88 16 Simple Form x xr = rx r 1 x sin x = cos x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x)
89 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x)
90 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] =
91 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x)
92 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x)
93 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] =
94 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x f (x)
95 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x f (x)
96 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x e f (x) = x f (x)
97 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x e f (x) = e f (x) x f (x) x f (x)
98 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x x ln x = 1 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x e f (x) = e f (x) x f (x) x f (x)
99 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x x ln x = 1 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x x e f (x) = e f (x) ln[ f (x)] = x f (x) x f (x)
100 16 Simple Form x xr = rx r 1 x x sin x = cos x cos x = sin x x tan x = sec2 x x ex = e x x ln x = 1 x General Form x [ f (x)]r = r[ f (x)] r 1 x f (x) x sin[ f (x)] = cos[ f (x)] x f (x) x cos[ f (x)] = sin[ f (x)] x f (x) x tan[ f (x)] = sec2 [ f (x)] x x e f (x) = e f (x) ln[ f (x)] = 1 f (x) x f (x) x f (x) x f (x)
101 17 Example Fin y x for the following: (1) y = sin(x 2 + 1) (2) y = e x3 +2 (3) y = ln(x 4 3x + 2) (4) y = e x5 +tan x 5 (5) y = ln[sin 2 (2x + 3)] (6) y = 1 (x 2 3e4x+1 + 3) 40 (7) y = e x+1 ln(x 2 + 1)
102 18 Example Fin y x for the following: (1) y = sin(x 2 + 1)
103 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution y x = x sin(x2 + 1)
104 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution y x = x sin(x2 + 1) = cos(x 2 + 1) x (x2 + 1)
105 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution y x = x sin(x2 + 1) = cos(x 2 + 1) = 2x cos(x 2 + 1) x (x2 + 1)
106 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution (2) y = e x3 +2 y x = x sin(x2 + 1) = cos(x 2 + 1) = 2x cos(x 2 + 1) x (x2 + 1)
107 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution (2) y = e x3 +2 Solution y x y x = x sin(x2 + 1) = cos(x 2 + 1) = 2x cos(x 2 + 1) = x ex3 +2 x (x2 + 1)
108 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution (2) y = e x3 +2 Solution y x y x = x sin(x2 + 1) = cos(x 2 + 1) = 2x cos(x 2 + 1) = x ex3 +2 = e x3 +2 x (x3 + 2) x (x2 + 1)
109 18 Example Fin y x for the following: (1) y = sin(x 2 + 1) Solution (2) y = e x3 +2 Solution y x y x = x sin(x2 + 1) = cos(x 2 + 1) = 2x cos(x 2 + 1) = x ex3 +2 = e x3 +2 = 3x 2 e x3 +2 x (x3 + 2) x (x2 + 1)
110 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2)
111 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2)
112 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2)
113 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2
114 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5
115 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5 Solution y x = x ex5 +tan x 5
116 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5 Solution y x = x ex5 +tan x 5 = e x5 +tan x 5 x (x5 + tan x 5 )
117 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5 Solution y x = x ex5 +tan x 5 = e x5 +tan x 5 x (x5 + tan x 5 ) = e x5 +tan x 5 (5x 4 + x tan (x5 ) o in your hea)
118 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5 Solution y x = x ex5 +tan x 5 = e x5 +tan x 5 x (x5 + tan x 5 ) = e x5 +tan x 5 (5x 4 + x tan (x5 ) o in your hea) = e x5 +tan x 5 (5x 4 + sec 2 x 5 x x5 )
119 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5 Solution y x = x ex5 +tan x 5 = e x5 +tan x 5 x (x5 + tan x 5 ) = e x5 +tan x 5 (5x 4 + x tan (x5 ) o in your hea) = e x5 +tan x 5 (5x 4 + sec 2 x 5 x x5 ) = e x5 +tan x 5 (5x 4 + 5x 4 sec 2 x 5 )
120 Example Fin y x for the following: 19 (3) y = ln(x 4 3x + 2) Solution y x = x ln(x4 3x + 2) = 1 x 4 3x + 2 x (x4 3x + 2) = 4x 3 3 x 4 3x + 2 (4) y = e x5 +tan x 5 Solution y x = x ex5 +tan x 5 = e x5 +tan x 5 x (x5 + tan x 5 ) = e x5 +tan x 5 (5x 4 + x tan (x5 ) o in your hea) = e x5 +tan x 5 (5x 4 + sec 2 x 5 x x5 ) = e x5 +tan x 5 (5x 4 + 5x 4 sec 2 x 5 ) = 5x 4 (1 + sec 2 x 5 )e x5 +tan x 5
20. Product rule, Quotient rule
20. Prouct rule, 20.1. Prouct rule Prouct rule, Prouct rule We have seen that the erivative of a sum is the sum of the erivatives: [f(x) + g(x)] = x x [f(x)] + x [(g(x)]. One might expect from this that
2 Integrating Both Sides
2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation
Using a table of derivatives
Using a table of derivatives In this unit we construct a Table of Derivatives of commonly occurring functions. This is done using the knowledge gained in previous units on differentiation from first principles.
correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:
Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that
Inverse Functions and Logarithms
Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka Inverse Functions and Logarithms DEFINITION: A function f is called a one-to-one function if it never takes on the same value twice; that
Calculus 1: Sample Questions, Final Exam, Solutions
Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.
Taylor and Maclaurin Series
Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions
Math 113 HW #7 Solutions
Math 3 HW #7 Solutions 35 0 Given find /dx by implicit differentiation y 5 + x 2 y 3 = + ye x2 Answer: Differentiating both sides with respect to x yields 5y 4 dx + 2xy3 + x 2 3y 2 ) dx = dx ex2 + y2x)e
f(x) = a x, h(5) = ( 1) 5 1 = 2 2 1
Exponential Functions an their Derivatives Exponential functions are functions of the form f(x) = a x, where a is a positive constant referre to as the base. The functions f(x) = x, g(x) = e x, an h(x)
Lecture 3: Derivatives and extremes of functions
Lecture 3: Derivatives and extremes of functions Lejla Batina Institute for Computing and Information Sciences Digital Security Version: spring 2011 Lejla Batina Version: spring 2011 Wiskunde 1 1 / 16
INTEGRATING FACTOR METHOD
Differential Equations INTEGRATING FACTOR METHOD Graham S McDonald A Tutorial Module for learning to solve 1st order linear differential equations Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk
Homework # 3 Solutions
Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8
The Derivative. Philippe B. Laval Kennesaw State University
The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition
Nonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
Differentiation and Integration
This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have
Microeconomic Theory: Basic Math Concepts
Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts
100. In general, we can define this as if b x = a then x = log b
Exponents and Logarithms Review 1. Solving exponential equations: Solve : a)8 x = 4! x! 3 b)3 x+1 + 9 x = 18 c)3x 3 = 1 3. Recall: Terminology of Logarithms If 10 x = 100 then of course, x =. However,
Second-Order Linear Differential Equations
Second-Order Linear Differential Equations A second-order linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1
Chapter 7 Outline Math 236 Spring 2001
Chapter 7 Outline Math 236 Spring 2001 Note 1: Be sure to read the Disclaimer on Chapter Outlines! I cannot be responsible for misfortunes that may happen to you if you do not. Note 2: Section 7.9 will
Section 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section.7 Rolle s Theorem and the Mean Value Theorem The two theorems which are at the heart of this section draw connections between the instantaneous rate
Inverse Trig Functions
Inverse Trig Functions c A Math Support Center Capsule February, 009 Introuction Just as trig functions arise in many applications, so o the inverse trig functions. What may be most surprising is that
CHAPTER 8: DIFFERENTIAL CALCULUS
CHAPTER 8: DIFFERENTIAL CALCULUS 1. Rules of Differentiation As we ave seen, calculating erivatives from first principles can be laborious an ifficult even for some relatively simple functions. It is clearly
Answers to the Practice Problems for Test 2
Answers to the Practice Problems for Test 2 Davi Murphy. Fin f (x) if it is known that x [f(2x)] = x2. By the chain rule, x [f(2x)] = f (2x) 2, so 2f (2x) = x 2. Hence f (2x) = x 2 /2, but the lefthan
INTERPOLATION. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y).
INTERPOLATION Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). As an example, consider defining and x 0 =0, x 1 = π 4, x
Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = e x. y = exp(x) if and only if x = ln(y)
Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = Last day, we saw that the function f(x) = ln x is one-to-one, with domain (, ) and range (, ). We can conclude that f(x) has an inverse function
Course outline, MA 113, Spring 2014 Part A, Functions and limits. 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems)
Course outline, MA 113, Spring 2014 Part A, Functions and limits 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems) Functions, domain and range Domain and range of rational and algebraic
Calculus. Contents. Paul Sutcliffe. Office: CM212a.
Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical
Integration by substitution
Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable
Sections 3.1/3.2: Introducing the Derivative/Rules of Differentiation
Sections 3.1/3.2: Introucing te Derivative/Rules of Differentiation 1 Tangent Line Before looking at te erivative, refer back to Section 2.1, looking at average velocity an instantaneous velocity. Here
Math 120 Final Exam Practice Problems, Form: A
Math 120 Final Exam Practice Problems, Form: A Name: While every attempt was made to be complete in the types of problems given below, we make no guarantees about the completeness of the problems. Specifically,
5.1 Derivatives and Graphs
5.1 Derivatives and Graphs What does f say about f? If f (x) > 0 on an interval, then f is INCREASING on that interval. If f (x) < 0 on an interval, then f is DECREASING on that interval. A function has
15.2. First-Order Linear Differential Equations. First-Order Linear Differential Equations Bernoulli Equations Applications
00 CHAPTER 5 Differential Equations SECTION 5. First-Orer Linear Differential Equations First-Orer Linear Differential Equations Bernoulli Equations Applications First-Orer Linear Differential Equations
Section 12.6: Directional Derivatives and the Gradient Vector
Section 26: Directional Derivatives and the Gradient Vector Recall that if f is a differentiable function of x and y and z = f(x, y), then the partial derivatives f x (x, y) and f y (x, y) give the rate
Introduction to Integration Part 1: Anti-Differentiation
Mathematics Learning Centre Introuction to Integration Part : Anti-Differentiation Mary Barnes c 999 University of Syney Contents For Reference. Table of erivatives......2 New notation.... 2 Introuction
Objective: Use calculator to comprehend transformations.
math111 (Bradford) Worksheet #1 Due Date: Objective: Use calculator to comprehend transformations. Here is a warm up for exploring manipulations of functions. specific formula for a function, say, Given
Homework #1 Solutions
MAT 303 Spring 203 Homework # Solutions Problems Section.:, 4, 6, 34, 40 Section.2:, 4, 8, 30, 42 Section.4:, 2, 3, 4, 8, 22, 24, 46... Verify that y = x 3 + 7 is a solution to y = 3x 2. Solution: From
Lectures 5-6: Taylor Series
Math 1d Instructor: Padraic Bartlett Lectures 5-: Taylor Series Weeks 5- Caltech 213 1 Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular,
Section 2.7 One-to-One Functions and Their Inverses
Section. One-to-One Functions and Their Inverses One-to-One Functions HORIZONTAL LINE TEST: A function is one-to-one if and only if no horizontal line intersects its graph more than once. EXAMPLES: 1.
x 2 if 2 x < 0 4 x if 2 x 6
Piecewise-defined Functions Example Consider the function f defined by x if x < 0 f (x) = x if 0 x < 4 x if x 6 Piecewise-defined Functions Example Consider the function f defined by x if x < 0 f (x) =
An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate.
Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set
FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA
FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA 1.1 Solve linear equations and equations that lead to linear equations. a) Solve the equation: 1 (x + 5) 4 = 1 (2x 1) 2 3 b) Solve the equation: 3x
Introduction to Differential Calculus. Christopher Thomas
Mathematics Learning Centre Introduction to Differential Calculus Christopher Thomas c 1997 University of Sydney Acknowledgements Some parts of this booklet appeared in a similar form in the booklet Review
Exponential Functions: Differentiation and Integration. The Natural Exponential Function
46_54.q //4 :59 PM Page 5 5 CHAPTER 5 Logarithmic, Eponential, an Other Transcenental Functions Section 5.4 f () = e f() = ln The inverse function of the natural logarithmic function is the natural eponential
Elementary Functions
Chapter Three Elementary Functions 31 Introduction Complex functions are, of course, quite easy to come by they are simply ordered pairs of real-valued functions of two variables We have, however, already
14.1. Basic Concepts of Integration. Introduction. Prerequisites. Learning Outcomes. Learning Style
Basic Concepts of Integration 14.1 Introduction When a function f(x) is known we can differentiate it to obtain its derivative df. The reverse dx process is to obtain the function f(x) from knowledge of
Notes and questions to aid A-level Mathematics revision
Notes and questions to aid A-level Mathematics revision Robert Bowles University College London October 4, 5 Introduction Introduction There are some students who find the first year s study at UCL and
TOPIC 4: DERIVATIVES
TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the
New Higher-Proposed Order-Combined Approach. Block 1. Lines 1.1 App. Vectors 1.4 EF. Quadratics 1.1 RC. Polynomials 1.1 RC
New Higher-Proposed Order-Combined Approach Block 1 Lines 1.1 App Vectors 1.4 EF Quadratics 1.1 RC Polynomials 1.1 RC Differentiation-but not optimisation 1.3 RC Block 2 Functions and graphs 1.3 EF Logs
Limits and Continuity
Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function
Techniques of Integration
CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration
12.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following:
Section 1.6 Logarithmic and Exponential Equations 811 1.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: Solve Quadratic Equations (Section
1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives
TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of
Differentiability of Exponential Functions
Differentiability of Exponential Functions Philip M. Anselone an John W. Lee Philip Anselone (panselone@actionnet.net) receive his Ph.D. from Oregon State in 1957. After a few years at Johns Hopkins an
Here the units used are radians and sin x = sin(x radians). Recall that sin x and cos x are defined and continuous everywhere and
Lecture 9 : Derivatives of Trigonometric Functions (Please review Trigonometry uner Algebra/Precalculus Review on the class webpage.) In this section we will look at the erivatives of the trigonometric
MATH 31B: MIDTERM 1 REVIEW. 1. Inverses. yx 3y = 1. x = 1 + 3y y 4( 1) + 32 = 1
MATH 3B: MIDTERM REVIEW JOE HUGHES. Inverses. Let f() = 3. Find the inverse g() for f. Solution: Setting y = ( 3) and solving for gives and g() = +3. y 3y = = + 3y y. Let f() = 4 + 3. Find a domain on
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form 1 Px d y dx dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise
CS 261 Fall 2011 Solutions to Assignment #4
CS 61 Fall 011 Solutions to Assignment #4 The following four algorithms are used to implement the bisection method, Newton s method, the secant method, and the method of false position, respectively. In
MATH 425, PRACTICE FINAL EXAM SOLUTIONS.
MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator
FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.
FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
Review of Matlab for Differential Equations. Lia Vas
Review of Matlab for Differential Equations Lia Vas 1. Basic arithmetic (Practice problems 1) 2. Solving equations with solve (Practice problems 2) 3. Representing functions 4. Graphics 5. Parametric Plots
y or f (x) to determine their nature.
Level C5 of challenge: D C5 Fining stationar points of cubic functions functions Mathematical goals Starting points Materials require Time neee To enable learners to: fin the stationar points of a cubic
Rules for Finding Derivatives
3 Rules for Fining Derivatives It is teious to compute a limit every time we nee to know the erivative of a function. Fortunately, we can evelop a small collection of examples an rules that allow us to
6.4 Logarithmic Equations and Inequalities
6.4 Logarithmic Equations and Inequalities 459 6.4 Logarithmic Equations and Inequalities In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies.
Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section 3.3 Differentiation of Polynomials an Rational Functions In tis section we begin te task of iscovering rules for ifferentiating various classes of
G.A. Pavliotis. Department of Mathematics. Imperial College London
EE1 MATHEMATICS NUMERICAL METHODS G.A. Pavliotis Department of Mathematics Imperial College London 1. Numerical solution of nonlinear equations (iterative processes). 2. Numerical evaluation of integrals.
SOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve
SOLUTIONS Problem. Find the critical points of the function f(x, y = 2x 3 3x 2 y 2x 2 3y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Partial derivatives
Find all of the real numbers x that satisfy the algebraic equation:
Appendix C: Factoring Algebraic Expressions Factoring algebraic equations is the reverse of expanding algebraic expressions discussed in Appendix B. Factoring algebraic equations can be a great help when
Section 1. Logarithms
Worksheet 2.7 Logarithms and Exponentials Section 1 Logarithms The mathematics of logarithms and exponentials occurs naturally in many branches of science. It is very important in solving problems related
1 Lecture: Integration of rational functions by decomposition
Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.
Chapter 7. Continuity
Chapter 7 Continuity There are many processes and eects that depends on certain set of variables in such a way that a small change in these variables acts as small change in the process. Changes of this
The Mathematics Diagnostic Test
The Mathematics iagnostic Test Mock Test and Further Information 010 In welcome week, students will be asked to sit a short test in order to determine the appropriate lecture course, tutorial group, whether
Math 229 Lecture Notes: Product and Quotient Rules Professor Richard Blecksmith richard@math.niu.edu
Mat 229 Lecture Notes: Prouct an Quotient Rules Professor Ricar Blecksmit ricar@mat.niu.eu 1. Time Out for Notation Upate It is awkwar to say te erivative of x n is nx n 1 Using te prime notation for erivatives,
Math 230.01, Fall 2012: HW 1 Solutions
Math 3., Fall : HW Solutions Problem (p.9 #). Suppose a wor is picke at ranom from this sentence. Fin: a) the chance the wor has at least letters; SOLUTION: All wors are equally likely to be chosen. The
REVIEW EXERCISES DAVID J LOWRY
REVIEW EXERCISES DAVID J LOWRY Contents 1. Introduction 1 2. Elementary Functions 1 2.1. Factoring and Solving Quadratics 1 2.2. Polynomial Inequalities 3 2.3. Rational Functions 4 2.4. Exponentials and
How To Understand The Theory Of Algebraic Functions
Homework 4 3.4,. Show that x x cos x x holds for x 0. Solution: Since cos x, multiply all three parts by x > 0, we get: x x cos x x, and since x 0 x x 0 ( x ) = 0, then by Sandwich theorem, we get: x 0
1.(6pts) Find symmetric equations of the line L passing through the point (2, 5, 1) and perpendicular to the plane x + 3y z = 9.
.(6pts Find symmetric equations of the line L passing through the point (, 5, and perpendicular to the plane x + 3y z = 9. (a x = y + 5 3 = z (b x (c (x = ( 5(y 3 = z + (d x (e (x + 3(y 3 (z = 9 = y 3
x 2 y 2 +3xy ] = d dx dx [10y] dy dx = 2xy2 +3y
![x 2 y 2 +3xy ] = d dx dx [10y] dy dx = 2xy2 +3y](/thumbs/40/21392919.jpg "x 2 y 2 +3xy ] = d dx dx [10y] dy dx = 2xy2 +3y")
MA7 - Calculus I for thelife Sciences Final Exam Solutions Spring -May-. Consider the function defined implicitly near (,) byx y +xy =y. (a) [7 points] Use implicit differentiation to find the derivative
6. Differentiating the exponential and logarithm functions
1 6. Differentiating te exponential and logaritm functions We wis to find and use derivatives for functions of te form f(x) = a x, were a is a constant. By far te most convenient suc function for tis purpose
3.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS. Copyright Cengage Learning. All rights reserved.
3.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS Copyright Cengage Learning. All rights reserved. What You Should Learn Recognize and evaluate logarithmic functions with base a. Graph logarithmic functions.
ALGEBRA REVIEW LEARNING SKILLS CENTER. Exponents & Radicals
ALGEBRA REVIEW LEARNING SKILLS CENTER The "Review Series in Algebra" is taught at the beginning of each quarter by the staff of the Learning Skills Center at UC Davis. This workshop is intended to be an
Simplify the rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression.
MAC 1105 Final Review Simplify the rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression. 1) 8x 2-49x + 6 x - 6 A) 1, x 6 B) 8x - 1, x 6 x -
Approximating functions by Taylor Polynomials.
Chapter 4 Approximating functions by Taylor Polynomials. 4.1 Linear Approximations We have already seen how to approximate a function using its tangent line. This was the key idea in Euler s method. If
MATH 381 HOMEWORK 2 SOLUTIONS
MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e
15.1. Exact Differential Equations. Exact First-Order Equations. Exact Differential Equations Integrating Factors
SECTION 5. Eact First-Order Equations 09 SECTION 5. Eact First-Order Equations Eact Differential Equations Integrating Factors Eact Differential Equations In Section 5.6, ou studied applications of differential
WORKBOOK. MATH 30. PRE-CALCULUS MATHEMATICS.
WORKBOOK. MATH 30. PRE-CALCULUS MATHEMATICS. DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE Contributor: U.N.Iyer Department of Mathematics and Computer Science, CP 315, Bronx Community College, University
6 Further differentiation and integration techniques
56 6 Further differentiation and integration techniques Here are three more rules for differentiation and two more integration techniques. 6.1 The product rule for differentiation Textbook: Section 2.7
Function Name Algebra. Parent Function. Characteristics. Harold s Parent Functions Cheat Sheet 28 December 2015
Harold s s Cheat Sheet 8 December 05 Algebra Constant Linear Identity f(x) c f(x) x Range: [c, c] Undefined (asymptote) Restrictions: c is a real number Ay + B 0 g(x) x Restrictions: m 0 General Fms: Ax
This makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5
1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy,
Section 4.5 Exponential and Logarithmic Equations
Section 4.5 Exponential and Logarithmic Equations Exponential Equations An exponential equation is one in which the variable occurs in the exponent. EXAMPLE: Solve the equation x = 7. Solution 1: We have
Average rate of change of y = f(x) with respect to x as x changes from a to a + h:
L15-1 Lecture 15: Section 3.4 Definition of the Derivative Recall the following from Lecture 14: For function y = f(x), the average rate of change of y with respect to x as x changes from a to b (on [a,
RAJALAKSHMI ENGINEERING COLLEGE MA 2161 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS PART A
RAJALAKSHMI ENGINEERING COLLEGE MA 26 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS. Solve (D 2 + D 2)y = 0. 2. Solve (D 2 + 6D + 9)y = 0. PART A 3. Solve (D 4 + 4)x = 0 where D = d dt 4. Find Particular Integral:
PRE-CALCULUS GRADE 12
PRE-CALCULUS GRADE 12 [C] Communication Trigonometry General Outcome: Develop trigonometric reasoning. A1. Demonstrate an understanding of angles in standard position, expressed in degrees and radians.
Find the length of the arc on a circle of radius r intercepted by a central angle θ. Round to two decimal places.
SECTION.1 Simplify. 1. 7π π. 5π 6 + π Find the measure of the angle in degrees between the hour hand and the minute hand of a clock at the time shown. Measure the angle in the clockwise direction.. 1:0.
Chapter 11. Techniques of Integration
Chapter Techniques of Integration Chapter 6 introduced the integral. There it was defined numerically, as the limit of approximating Riemann sums. Evaluating integrals by applying this basic definition
x(x + 5) x 2 25 (x + 5)(x 5) = x 6(x 4) x ( x 4) + 3
CORE 4 Summary Notes Rational Expressions Factorise all expressions where possible Cancel any factors common to the numerator and denominator x + 5x x(x + 5) x 5 (x + 5)(x 5) x x 5 To add or subtract -
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 14 10/27/2008 MOMENT GENERATING FUNCTIONS
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 14 10/27/2008 MOMENT GENERATING FUNCTIONS Contents 1. Moment generating functions 2. Sum of a ranom number of ranom variables 3. Transforms
Techniques of Integration
8 Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it
x), etc. In general, we have
BASIC CALCULUS REFRESHER. Introduction. Ismor Fischer, Ph.D. Dept. of Statistics UW-Madison This is a very condensed and simplified version of basic calculus, which is a prerequisite for many courses in
Analysis of errors in derivatives of trigonometric functions
Siyepu International Journal of STEM Education (2015) 2:16 DOI 10.1186/s40594-015-0029-5 SHORT REPORT Open Access Analysis of errors in derivatives of trigonometric functions Sibawu Witness Siyepu Abstract
Lagrangian and Hamiltonian Mechanics
Lagrangian an Hamiltonian Mechanics D.G. Simpson, Ph.D. Department of Physical Sciences an Engineering Prince George s Community College December 5, 007 Introuction In this course we have been stuying