To differentiate logarithmic functions with bases other than e, use


 Eleanore Lyons
 2 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
M147 Practice Problems for Exam 2
M47 Practice Problems for Exam Exam will cover sections 4., 4.4, 4.5, 4.6, 4.7, 4.8, 5., an 5.. Calculators will not be allowe on the exam. The first ten problems on the exam will be multiple choice. Work
More information20. 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
More informationMATH 125: LAST LECTURE
MATH 5: LAST LECTURE FALL 9. Differential Equations A ifferential equation is an equation involving an unknown function an it s erivatives. To solve a ifferential equation means to fin a function that
More informationLecture 13: Differentiation Derivatives of Trigonometric Functions
Lecture 13: Differentiation Derivatives of Trigonometric Functions Derivatives of the Basic Trigonometric Functions Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the
More information2 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
More informationSolutions to modified 2 nd Midterm
Math 125 Solutions to moifie 2 n Miterm 1. For each of the functions f(x) given below, fin f (x)). (a) 4 points f(x) = x 5 + 5x 4 + 4x 2 + 9 Solution: f (x) = 5x 4 + 20x 3 + 8x (b) 4 points f(x) = x 8
More informationProof of the Power Rule for Positive Integer Powers
Te Power Rule A function of te form f (x) = x r, were r is any real number, is a power function. From our previous work we know tat x x 2 x x x x 3 3 x x In te first two cases, te power r is a positive
More informationLecture 17: Implicit differentiation
Lecture 7: Implicit ifferentiation Nathan Pflueger 8 October 203 Introuction Toay we iscuss a technique calle implicit ifferentiation, which provies a quicker an easier way to compute many erivatives we
More informationUsing 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.
More informationD f = (2, ) (x + 1)(x 3) (b) g(x) = x 1 solution: We need the thing inside the root to be greater than or equal to 0. So we set up a sign table.
. Find the domains of the following functions: (a) f(x) = ln(x ) We need x > 0, or x >. Thus D f = (, ) (x + )(x 3) (b) g(x) = x We need the thing inside the root to be greater than or equal to 0. So we
More informationcorrectchoice 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
More informationInverse Functions and Logarithms
Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka Inverse Functions and Logarithms DEFINITION: A function f is called a onetoone function if it never takes on the same value twice; that
More informationCalculus 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.
More informationMath 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
More informationVersion PREVIEW HW 04 hoffman (57225) 1. Consequently,
Version PREVIEW HW 04 hoffman (57225) 1 This printout shoul have 12 questions Multiplechoice questions may continue on the next column or page fin all choices before answering CalCa02b 001 100 points
More informationSection 3.1 Worksheet NAME. f(x + h) f(x)
MATH 1170 Section 3.1 Worksheet NAME Recall that we have efine the erivative of f to be f (x) = lim h 0 f(x + h) f(x) h Recall also that the erivative of a function, f (x), is the slope f s tangent line
More informationf(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)
More informationTaylor 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
More informationDerivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x):
Derivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x): f f(x + h) f(x) (x) = lim h 0 h (for all x for which f is differentiable/ the limit exists) Property:if
More informationMAT1A01: Differentiation of Polynomials & Exponential Functions + the Product & Quotient Rules
MAT1A01: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 17 April 2013 Reminer Mats Learning Centre: CRing 512 My office: CRing 533A (Stats Dept corrior)
More informationLecture 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
More information2 HYPERBOLIC FUNCTIONS
HYPERBOLIC FUNCTIONS Chapter Hyperbolic Functions Objectives After stuying this chapter you shoul unerstan what is meant by a hyperbolic function; be able to fin erivatives an integrals of hyperbolic functions;
More informationINTEGRATING 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
More informationChapter Usual types of questions Tips What can go ugly 1 Algebraic Fractions
C3 Cheat Sheet Last Updated: 9 th Nov 2015 Chapter Usual types of questions Tips What can go ugly 1 Algebraic Fractions Almost always adding or subtracting fractions. Factorise everything in each fraction
More information19.2. First Order Differential Equations. Introduction. Prerequisites. Learning Outcomes
First Orer Differential Equations 19.2 Introuction Separation of variables is a technique commonly use to solve first orer orinary ifferential equations. It is socalle because we rearrange the equation
More informationnparameter families of curves
1 nparameter families of curves For purposes of this iscussion, a curve will mean any equation involving x, y, an no other variables. Some examples of curves are x 2 + (y 3) 2 = 9 circle with raius 3,
More informationf(x) = undefined otherwise We have Domain(f) = [ π, π ] and Range(f) = [ 1, 1].
Lecture 6 : Inverse Trigonometric Functions Inverse Sine Function (arcsin = sin ) The trigonometric function sin is not onetoone functions, hence in orer to create an inverse, we must restrict its omain.
More informationHomework # 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
More informationInverse Trig Functions
Inverse Trig Functions Trig functions are not onetoone, so we can not formally get an inverse. To efine the notion of inverse trig functions we restrict the omains to obtain onetoone functions: [ Restrict
More informationThe 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
More informationDefinition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =
Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a
More informationLearning Objectives for Math 165
Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given
More information5 Indefinite integral
5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse
More informationA: TABLE OF BASIC DERIVATIVES
A: TABLE OF BASIC DERIVATIVES Let u = u(x) be a ifferentiable function of the inepenent variable x, that is u (x) exists. (A) The Power Rule : Examples : x {un } = nu n. u x {(x3 + 4x + ) 3/4 } = 3 4 (x3
More informationDifferentiation 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
More informationNonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve secondorder nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
More informationThe Inverse Trigonometric Functions
The Inverse Trigonometric Functions These notes amplify on the book s treatment of inverse trigonometric functions an supply some neee practice problems. Please see pages 543 544 for the graphs of sin
More informationMicroeconomic 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
More information100. 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,
More informationChapter 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
More informationSept 20, 2011 MATH 140: Calculus I Tutorial 2. ln(x 2 1) = 3 x 2 1 = e 3 x = e 3 + 1
Sept, MATH 4: Calculus I Tutorial Solving Quadratics, Dividing Polynomials Problem Solve for x: ln(x ) =. ln(x ) = x = e x = e + Problem Solve for x: e x e x + =. Let y = e x. Then we have a quadratic
More informationSecondOrder Linear Differential Equations
SecondOrder Linear Differential Equations A secondorder 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
More informationCalculus. 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
More informationPractice Problems for Midterm 2
Practice Problems for Midterm () For each of the following, find and sketch the domain, find the range (unless otherwise indicated), and evaluate the function at the given point P : (a) f(x, y) = + 4 y,
More informationCHAPTER 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
More informationInverse 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
More information1 Chapter Chapter Chapter Chapter 4 1
Contents 1 Chapter 1 1 2 Chapter 2 1 3 Chapter 3 1 4 Chapter 4 1 5 Applications of Integrals 2 5.1 Area and Definite Integrals............................ 2 5.1.1 Area of a Region Between Two Curves.................
More informationAnswers 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
More informationLecture 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 onetoone, with domain (, ) and range (, ). We can conclude that f(x) has an inverse function
More informationSection 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
More informationk=1 k2, and therefore f(m + 1) = f(m) + (m + 1) 2 =
Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 +2 2 + +n 2 = 1 n(n+1)(2n+1) for all n N. 6 Put f(n) = n(n + 1)(2n + 1)/6. Then f(1) = 1, i.e the theorem
More informationInverse Trig Functions
Inverse Trig Functions Previously in Math 30 DEFINITION 24 A function g is the inverse of the function f if g( f ()) = for all in the omain of f 2 f (g()) = for all in the omain of g In this situation
More informationINTERPOLATION. 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
More informationIntegration 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
More informationCourse outline, MA 113, Spring 2014 Part A, Functions and limits. 1.1 1.2 Functions, domain and ranges, A1.11.2Review (9 problems)
Course outline, MA 113, Spring 2014 Part A, Functions and limits 1.1 1.2 Functions, domain and ranges, A1.11.2Review (9 problems) Functions, domain and range Domain and range of rational and algebraic
More informationSections 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
More informationMath 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,
More informationLimit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)
SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f
More informationPreCalculus Review Lesson 1 Polynomials and Rational Functions
If a and b are real numbers and a < b, then PreCalculus Review Lesson 1 Polynomials and Rational Functions For any real number c, a + c < b + c. For any real numbers c and d, if c < d, then a + c < b
More information5.1 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
More informationSection 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
More information15.2. FirstOrder Linear Differential Equations. FirstOrder Linear Differential Equations Bernoulli Equations Applications
00 CHAPTER 5 Differential Equations SECTION 5. FirstOrer Linear Differential Equations FirstOrer Linear Differential Equations Bernoulli Equations Applications FirstOrer Linear Differential Equations
More informationIntroduction to Integration Part 1: AntiDifferentiation
Mathematics Learning Centre Introuction to Integration Part : AntiDifferentiation Mary Barnes c 999 University of Syney Contents For Reference. Table of erivatives......2 New notation.... 2 Introuction
More informationItems related to expected use of graphing technology appear in bold italics.
 1  Items related to expected use of graphing technology appear in bold italics. Investigating the Graphs of Polynomial Functions determine, through investigation, using graphing calculators or graphing
More informationHomework #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
More informationMATH 2300 review problems for Exam 3 ANSWERS
MATH 300 review problems for Exam 3 ANSWERS. Check whether the following series converge or diverge. In each case, justify your answer by either computing the sum or by by showing which convergence test
More informationLecture 5 : Continuous Functions Definition 1 We say the function f is continuous at a number a if
Lecture 5 : Continuous Functions Definition We say the function f is continuous at a number a if f(x) = f(a). (i.e. we can make the value of f(x) as close as we like to f(a) by taking x sufficiently close
More informationPartial differentiation
Chapter 1 Partial differentiation Example 1.1 What is the maximal domain of the real function g defined b g(x) = x 2 + 3x + 2? : The ke point is that the square root onl gives a real result if the argument
More informationLectures 56: 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,
More informationSOME NOTES ON THE HYPERBOLIC TRIG FUNCTIONS SINH AND COSH
SOME NOTES ON THE HYPERBOLIC TRIG FUNCTIONS SINH AND COSH Basic Definitions In homework set # one of the questions involves basic unerstaning of the hyperbolic functions sinh an cosh. We will use this
More information1 Mathematical Induction
Extra Credit Homework Problems Note: these problems are of varying difficulty, so you might want to assign different point values for the different problems. I have suggested the point values each problem
More informationx 2 if 2 x < 0 4 x if 2 x 6
Piecewisedefined Functions Example Consider the function f defined by x if x < 0 f (x) = x if 0 x < 4 x if x 6 Piecewisedefined Functions Example Consider the function f defined by x if x < 0 f (x) =
More informationAn 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
More informationSection 2.7 OnetoOne Functions and Their Inverses
Section. OnetoOne Functions and Their Inverses OnetoOne Functions HORIZONTAL LINE TEST: A function is onetoone if and only if no horizontal line intersects its graph more than once. EXAMPLES: 1.
More information14.1. 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
More informationObjective: 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
More informationFINAL 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
More informationIntroduction 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
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Math 110 Review for Final Examination 2012 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Match the equation to the correct graph. 1) y = 
More informationExponential 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
More informationElementary Functions
Chapter Three Elementary Functions 31 Introduction Complex functions are, of course, quite easy to come by they are simply ordered pairs of realvalued functions of two variables We have, however, already
More informationHyperbolic functions (CheatSheet)
Hyperbolic functions (CheatSheet) 1 Intro For historical reasons hyperbolic functions have little or no room at all in the syllabus of a calculus course, but as a matter of fact they have the same ignity
More informationNotes and questions to aid Alevel Mathematics revision
Notes and questions to aid Alevel 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
More informationTrigonometric Limits. more examples of limits
Trigonometric Limits more examples of its Substitution Theorem for Trigonometric Functions laws for evaluating its Theorem A. For each point c in function s domain: x c x c x c sin x = sin c, tan x = tan
More informationTOPIC 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
More informationTaylor Polynomials and Taylor Series Math 126
Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will
More information1 Derivatives of Piecewise Defined Functions
MATH 1010E University Matematics Lecture Notes (week 4) Martin Li 1 Derivatives of Piecewise Define Functions For piecewise efine functions, we often ave to be very careful in computing te erivatives.
More informationInverse Trigonometric Functions
Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka Inverse Trigonometric Functions DEFINITION: The inverse sine function, denoted by sin x or arcsin x), is defined to be the inverse of the
More informationMATHEMATICS (CLASSES XI XII)
MATHEMATICS (CLASSES XI XII) General Guidelines (i) All concepts/identities must be illustrated by situational examples. (ii) The language of word problems must be clear, simple and unambiguous. (iii)
More informationNew HigherProposed OrderCombined Approach. Block 1. Lines 1.1 App. Vectors 1.4 EF. Quadratics 1.1 RC. Polynomials 1.1 RC
New HigherProposed OrderCombined Approach Block 1 Lines 1.1 App Vectors 1.4 EF Quadratics 1.1 RC Polynomials 1.1 RC Differentiationbut not optimisation 1.3 RC Block 2 Functions and graphs 1.3 EF Logs
More information6.6 The Inverse Trigonometric Functions. Outline
6.6 The Inverse Trigonometric Functions Tom Lewis Fall Semester 2015 Outline The inverse sine function The inverse cosine function The inverse tangent function The other inverse trig functions Miscellaneous
More informationTechniques of Integration
CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an artform than a collection of algorithms. Many problems in applied mathematics involve the integration
More informationUsing implicit di erentiation for good: Inverse functions.
Using implicit di erentiation for good: Inverse functions. Warmup: Calculate dy dx if 1. e y = xy Take d dx So of both sides to find ey dy dx = x dy dx + y. y = e y dy dx x dy dx = dy dx (ey x), implying
More informationLimits 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
More informationFinding Antiderivatives and Evaluating Integrals
Chapter 5 Finding Antiderivatives and Evaluating Integrals 5. Constructing Accurate Graphs of Antiderivatives Motivating Questions In this section, we strive to understand the ideas generated by the following
More informationSome Notes on Taylor Polynomials and Taylor Series
Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited
More information12.6 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
More informationTest # 2 Review. function y sin6x such that dx. per second. Find dy. f(x) 3x 2 6x 8 using the limiting process. dt = 2 centimeters. dt when x 7.
Name: Class: Date: ID: A Test # 2 Review Short Answer 1. Find the slope m of the line tangent to the graph of the function g( x) 9 x 2 at the point 4, 7ˆ. 2. A man 6 feet tall walks at a rate of 2 ft per
More informationx 2 + y 2 = 25 and try to solve for y in terms of x, we get 2 new equations y = 25 x 2 and y = 25 x 2.
Lecture : Implicit differentiation For more on the graphs of functions vs. the graphs of general equations see Graphs of Functions under Algebra/Precalculus Review on the class webpage. For more on graphing
More informationHere 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
More information