To differentiate logarithmic functions with bases other than e, use


 Eleanore Lyons
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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
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