Lesson 2 Derivatives and Integrals Involving The

LESSON 2 DERIVATIVES AND INTEGRALS INVOLVING THE NATURAL AND GENERAL EXPONENTIAL FUNCTIONSDefinition The exponential function with base b is the function defined by f ( x )  b x , where b  0 and b  1.Recall the following information about exponential functions:1. The domain of f ( x )  b x is the set of real numbers. That is, the domain of f ( x )  b x is (  ,  ) . 2. The range of f ( x )  b x is the set of positive real numbers. That is, the range of f ( x )  b x is (0,  ) . 3. The exponential function f ( x )  b x and the logarithmic function g( x )  log b x are inverses of one another. We will study the calculus of logarithmic functions in Lesson 3.Definition The natural exponential function is the exponential function whose base is the irrational number e. Thus, the natural exponential function is the function defined by f ( x )  e x , where e  2.71828.... .Notation: Sometimes, e x is written exp( x ) . e t  1 Theorem lim  1 t  0 tTheorem If f ( x )  e x , then f ( x )  e x . e x  h  e x e x e h  e x e x ( e h  1) Proof f ( x )  lim = lim = lim = h  0 h h  0 h h  0 h e h  1 e x lim = e x (1)  e x . h  0 hCopyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 dy du dy Theorem If y  e u , where u  g( x ) , then  e u or  e g( x ) g( x ) . dx dx dx x x Proof Use the Chain Rule and the fact that D x e  e .Theorem If f ( x )  b x , then f ( x )  b x ln b .Recall: ln b is the natural logarithm of b. That is, ln b  log e b . That is, it is the logarithm of b base e, where e  2.71828.... .Proof Since the natural logarithm function is the inverse of the natural exponential function, then we have that e ln x  x for all x in the domain of the natural logarithm function, which is the set of positive real numbers. Since b x is a x positive real number for all real numbers x, then we have that e ln b  b x for all real numbers x. Since ln b x  x ln b , then b x  e x ln b for all real numbers x. Thus, f ( x )  b x  e x ln b for all x in the domain of the exponential function. x ln b x ln b x ln b Since f ( x )  e , then f ( x )  e D x ( x ln b) = e ln b . Since b x  e x ln b , then f ( x )  b x ln b . dy du Theorem If y  b u , where u  g( x ) , then  b u (ln b) or dx dx dy  b g( x ) (ln b) g( x ) . dx x x Proof Use the Chain Rule and the fact that D x b  b ln b .Examples Differentiate the following functions.1. f ( x )  e 5 x5 x 5 x 5 x f ( x )  e D x (5 x ) = e 5 = 5eCopyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Answer: f ( x )  5e 5 x2 2. y  4 3  x2 3  x 2 3  x 2 y  4 (ln 4) D x (3  x ) = 4 (ln 4)(  2 x ) =2 2  x 4 3  x ( 2ln 4) =  x 4 3  x (ln 16)2 2 Answer: y   x 4 3  x ( 2ln 4) or y   x 4 3  x (ln 16)2 NOTE: There is another way to work this problem. Since 4 3  x = 2 2 2 4 3 4  x = 64  4  x , then y  64  4  x . Thus, 2 2 y  64  4  x (ln 4)(  2 x ) =  128 x 4  x ln 4 .3 3. g( w)  e 7 / 6w3  7  e 7 / 6w  exp  NOTE:  3   6 w 7 / 6w3  7  3  7 / 6w3  7  4  7  4 7 / 6w3 g( w)  e D w  w  = e   w  =  w e  6   2  27 3 Answer: g( w)   w  4 e 7 / 6w 24. s(t )  3 t tan 3 t3 t tan t 3 s(t )  3 (ln 3) D t (t tan t ) = Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 t tan 3 t  1  3 ln 3  tan 3 t  t (sec 2 3 t ) t  2 / 3  =  3  t tan 3 t  1  3 ln 3  tan 3 t  3 t sec 2 3 t  =  3 3 t tan 3 t ln 3 3tan 3 t  3 t sec 2 3 t  = 33 3 t tan t  1 ln 3 3tan 3 t  3 t sec 2 3 t 3 Answer: s(t )  3 t tan t  1 ln 3 3tan 3 t  3 t sec 2 3 t  or3 s(t )  3 t tan t ln 3 3 3tan 3 t  3 t sec 2 3 t 4 2 4 5. h( x )  e 18  3x  x4 18  3x 2  x 4 2 4 1/ 4 h( x )  e D x (18  3 x  x ) =4 2 4 1 e 18  3x  x (18  3 x 2  x 4 )  3/ 4 (6 x  4 x 3 ) = 44 2 4 4 2 4 2 x (3  2 x 2 )e 18  3 x  x x (3  2 x 2 )e 18  3 x  x = 4(18  3 x 2  x 4 ) 3/ 4 2(18  3 x 2  x 4 ) 3/ 44 2 4 x (3  2 x 2 )e 18  3 x  x Answer: h( x )  2(18  3 x 2  x 4 ) 3/ 46. y  cos( 2    )Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 dy   [sin( 2    ) ] D ( 2    ) =  [sin( 2    )]( 2  ln 2  1) d = (1  2  ln 2)sin( 2    ) dy Answer:  (1  2  ln 2)sin ( 2    ) d4 7. y  (3e  u  cscu3 )6 dy 4 4  6(3e  u  cscu3 )5 D (3e  u  cscu 3 ) = du u4 4 6(3e  u  cscu 3 )5 [3e  u (  4u 3 )  (  cscu3 cot u3 )3u 2 ] =4 4 6(3e  u  cscu 3 )5 [  3u 2 ( 4u e  u  cscu3 cot u3 )] =4 4  18u 2 (3e  u  cscu3 )5 ( 4u e  u  cscu 3 cot u3 ) dy 4 4 Answer:   18u 2 (3e  u  cscu 3 )5 ( 4u e  u  cscu3 cot u 3 ) du6  x  sin e2 x 8. f ( x )  9 x 2  x 3  x  1  1/ 2  2 x 2 x  2 3  x 2 x 2  6 (ln 6)  x   (cose )e 2 (9 x  x )  (6  sin e )(18 x  3x )  2  f ( x )    (9 x 2  x 3 )23 5 9. e x y  e 3x cot 8 y  x 2e yCopyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 We will use implicit differentiation in order to find the derivative of y with dy respect to x. That is , to find  y . dx3 5 3 5 x y x y 3 5 x 3 y 5 2 5 3 4 NOTE: D x e  e D x ( x y ) = e (3x y  x 5 y y)3x 3x 3x D x ( e cot 8 y )  ( D x e )cot 8 y  e D x cot 8 y =3e 3x cot 8 y  e 3x (  csc 2 8 y )8 y2 y 2 y 2 y y 2 y D x ( x e ) = ( D x x )e  x D x e = 2 x e  x e y3 5 Thus, e x y  e 3x cot 8 y  x 2e y  3 5 e x y (3 x 2 y 5  x 3 5 y 4 y)  (3e 3x cot 8 y  e 3x (  csc 2 8 y )8 y) =2 x e y  x 2e y y 3 5 3 5 3x 2 y 5e x y  5 x 3 y 4 ye x y  3e 3x cot 8 y  8 ye 3x csc 2 8 y =2 x e y  x 2 ye y 3 5 5 x 3 y 4 ye x y  8 ye 3x csc 2 8 y  x 2 ye y =3 5 2 x e y  3 x 2 y 5e x y  3e 3x cot 8 y 3 5 y(5 x 3 y 4 e x y  8e 3x csc 2 8 y  x 2 e y ) =3 5 2 x e y  3 x 2 y 5e x y  3e 3x cot 8 y 3 5 2 x e y  3x 2 y 5e x y  3e 3x cot 8 y y  3 5 5 x 3 y 4 e x y  8e 3x csc 2 8 y  x 2 e yCopyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 3 5 2 x e y  3x 2 y 5e x y  3e 3x cot 8 y Answer: y  3 5 5 x 3 y 4 e x y  8e 3x csc 2 8 y  x 2 e yExamples Determine the interval(s) on which the following functions are increasing and decreasing. Then find the local maximum(s) and local minimum(s) of the function.4 2 4 1. h( x )  e 18  3x  xBecause of the even (4) index of the radical, we need to find when the expression 18  3 x 2  x 4 is greater than or equal to zero in order to determine the domain of the function h since the fourth root of a negative number is not a real number. Noting that the expression 18  3x 2  x 4 is quadratic in x 2 , we can use the quadratic formula to solve the equation 18  3 x 2  x 4  0 if the expression 18  3x 2  x 4 does not factor. Since 18  3x 2  x 4 = (3  x 2 )(6  x 2 ) , then 18  3 x 2  x 4  0  (3  x 2 )(6  x 2 )  0  x 2   3 or x 2  6 . The two solutions to the equation x 2   3 are complex, and the two real solutions to the equation x 2  6 are x   6 .Sign of ( 3  x 2 )( 6  x 2 ) :  +     6 6Thus, 18  3x 2  x 4  0 when x is a number in the interval [  6 , 6 ] . Thus, the domain of the function h is the interval [  6 , 6 ] . Recall, that the domain of the derivative of a function is a subset of the domain of the function. Since the domain of the function h is the interval [  6 , 6 ] , then the domain of the function h is a subset of [  6 , 6 ] .Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 4 2 4 x (3  2 x 2 )e 18  3 x  x From Example 5 above, we have that h( x )  . 2(18  3 x 2  x 4 ) 3/ 4 Thus, the domain of h is (  6 , 6 ) .4 2 4 Since e 18  3x  x  0 , then h( x )  0  x (3  2 x 2 )  0  3 x  0 , x   . Note that h is undefined when 18  3x 2  x 4  0  2 x   6 . These numbers are not in the domain of h.4 2 4 NOTE: We know that the exponential e 18  3x  x is positive on its domain of definition. Since we showed that the expression 18  3x 2  x 4 is positive on the interval (  6 , 6 ) . Thus, the expression (18  3x 2  x 4 ) 3/ 4 is positive on this interval. Thus, the factors2 4 18  3 x 2  x 4 4 2 4 2 4 3/ 4 x (3  2 x )e e 18  3x  x and (18  3x  x ) in h( x )  2(18  3x 2  x 4 ) 3/ 4 are positive on the interval (  6 , 6 ) . Thus, the only factors that affect the sign of h are x and 3  2 x 2 .Sign of h( x ) : +  +       3 3  6  0 6 2 2 3   3       Since h ( x )  0 on the set   6 ,     0,  , then h is  2   2  increasing on these intervals. 3   3       Since h ( x )  0 on the set  , 0    , 6  , then h is  2   2  decreasing on these intervals.Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 3 3 There are local maximums occurring when x   and x  . The 2 2  3   3      local maximums are h    and h   . There is a local minimum  2   2  occurring when x  0 and the local minimum is h(0) .4 2 4 4 2 4 Since h( x )  e 18  3x  x , then h(  x )  e 18  3(  x )  (  x ) =4 2 4 e 18  3x  x  h( x ) for all x in the domain of h, then h is an even  3   3      function and h    = h   .  2   2  3   9 9    1 1   h    exp 4 18    exp 4 9 2        =     =  2   2 4    2 4    8 2 1     9    81  exp 4 9    exp 4 9  exp 4      =     =   =   4 4 4     4    4  3   3 4 4   3 4 4  exp  = exp  = exp  = 3 4 4 / 2 = 3 2 / 2 since  4   4    e e  4   16   2 4 4  4 2 2  2 2 / 4  2 1/ 2  2 h(0)  e 4 18 3   3      Answer: Increasing:   6 ,     0,   2   2  3   3      Decreasing:  , 0    , 6   2   2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Local Maximum: e 3 4 4 / 2 or e 3 2 / 2Local Minimum: e 4 183 2. f ( x )  x 2 e xNOTE: The domain of f is the set of real numbers.Using the Product Rule, we have that3 3 3 f ( x )  2 x e x  x 2 e x 3 x 2 = x e x ( 2  3 x 3 )2 2 f ( x )  0  x  0, x  3    3 3 3Sign of f ( x ) : +  +   2  3 0 3 2    3  Since f ( x )  0 on the set    ,     0,  , then f is  3  increasing on these intervals. 2    3  Since f ( x )  0 on the interval   , 0  , then f is decreasing on this  3  interval.2 There is a local maximum occurring when x   3 and the local 3  2   3  maximum is f    . There is a local minimum occurring when x  0  3  and the local minimum is f (0) .Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 3  2  4 12 2 x  3  3  2 / 3 3  2 / 3 Since f ( x )  x e , then f     e  e =  3  9 27 3 12 e  2 / 3 and f (0)  0. 3 2   3  Answer: Increasing:    ,     0,    3  2   3  Decreasing:   , 0   3 3 12 3 12 Local Maximum: e  2 / 3 or 3 3e 2 / 3Local Minimum: 03. g( x )  x 3 e 4 xNOTE: The domain of g is the set of real numbers.Using the Product Rule, we have that g( x )  3 x 2 e 4 x  x 3 e 4 x 4 = x 2 e 4 x (3  4 x )3 g( x )  0  x  0, x   4Sign of g( x ) :  + +   3  0 4 3  Thus, the function g is increasing on the interval   ,   .  4  Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860  3  Thus, the function g is decreasing on the interval    ,   .  4 3 There is a local minimum occurring when x   and the local minimum 4  3  is g    .  4  3  27 27 Since g( x )  x 3 e 4 x , then g      e  3   .  4  64 64 e 3 3  Answer: Increasing:   ,    4  3  Decreasing:    ,    4 Local Maximum: None27 27 Local Minimum:  e  3 or  64 64e 33 Example Determine the interval(s) on which the function f ( x )  x 2 e x is concave upward and concave downward.3 From Example 2 above, we have that f ( x )  x e x ( 2  3 x 3 ) = 3 ( 2 x  3x 4 )e x . Using the Product Rule again, we have that3 3 f ( x )  ( 2  12 x 3 )e x  ( 2 x  3 x 4 )e x 3 x 2 =3 3 e x [2  12 x 3  3 x 2 ( 2 x  3 x 4 )] = e x ( 2  12 x 3  6 x 3  9 x 6 ) =3 e x (9 x 6  18 x 3  2) .Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Noting that the expression 9 x 6  18 x 3  2 is quadratic in x 3 , we can use the quadratic formula to solve the equation 9 x 6  18 x 3  2  0 if the expression 9 x 6  18 x 3  2 does not factor. Since it does not factor, we will use the quadratic formula. b  b2  4a c  18  1818  4(9)( 2) x 3  = = 2a 18 18  1818  4(18)  18  18[18  4]  18  18(14) = = = 18 18 18 18  36( 7)  18  6 7 6(  3  7 )  3  7 = = = . Thus, 18 18 18 3 3  7  3  7  3  7 x 3  or x 3  . The equation x 3  has 3 3 3 three solutions. Two are complex and one is real. The one real solution can be  3  7  3  7 obtained by using cube roots. Thus, x 3   x  3 . 3 3  3  7  3  7 Similarly, x 3   x  3 . 3 3Sign of f ( x ) : +  +    3  7  3  7 3 3 3 3  3  7    3  7  Since f ( x )  0 on the set    , 3    3 ,   ,  3   3      then f is concave upward on these intervals.Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860   3  7  3  7  Since f ( x )  0 on the interval  3 , 3  , then f is  3 3    concave downward on this interval. 3  7 NOTE: There is an inflection point occurring when x  3 and the 3   3  7  y-coordinate of this point is f  3  . There is an inflection point  3     3  7 occurring when x  3 and the y-coordinate of this point is 3   3  7  f  3  .  3     3  7    3  7  Answer: Concave Upward:    , 3    3 ,    3   3       3  7  3  7  Concave Downward:  3 , 3   3 3   Examples Determine the interval(s) on which the following functions are increasing and decreasing, the interval(s) on which they are concave upward and concave downward. Also, find their local maximum(s), local minimum(s), and inflection point(s). x 1. f ( x )  e 3x e 3x  x e 3x 3 If we use the Quotient Rule, we obtain that f ( x )  = ( e 3x )2Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 e 3x (1  3x ) 1  3 x = . e 6 x e 3xIf we rewrite the function as f ( x )  x e  3x and use the Product Rule, we obtain that f ( x )  e  3x  x e  3x (  3) = e  3x (1  3x ) .Using the Product Rule again on f ( x )  (1  3 x )e  3 x , we obtain that f ( x )   3e  3x  (1  3x )e  3x (  3) =  3e  3x (1  1  3 x ) = 3e  3x ( 2  3 x )Sign of f ( x) : +   13 1  The function f is increasing on the interval   ,  and is decreasing on  3   1  the interval  ,   .  3 1 The function f has a local maximum occurring when x  and the local 3  1  1 1 maximum is f    e  1  .  3  3 3eSign of f ( x ) :  +  23Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860  2  The function f is concave upward on the interval  ,   and is concave  3   2  downward on the interval   ,  .  3 2 The function f has an inflection point at x  and the y-coordinate of the 3  2  2 2 point is f    e  2  .  3  3 3e 2 1   1  Answer: Increasing:   ,  Decreasing:  ,    3   3 1 Local Maximum: Local Minimum: None 3e 2   2  CU:  ,   CD:   ,   3   3  2 2   ,  Inflection Point:  2   3 3e 2. g( x )  x8 xUsing the Product Rule, we obtain that g( x )  8 x  x8 x ln 8 =8 x (1  x ln 8) .Using the Product Rule again on g( x )  (1  x ln 8)8 x , we obtain that g( x )  (ln 8)8 x  (1  x ln 8)8 x ln 8 = 8 x (ln 8)(1  1  x ln 8) =8 x (ln 8)( 2  x ln 8)Sign of g( x ) :  +Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860  1  ln 8 1  The function g is increasing on the interval   ,   and is decreasing  ln 8   1  on the interval   ,   .  ln 8 1 The function g has a local minimum occurring when x   and the ln 8   1 x x ln 8 x local minimum is g    . Since g( x )  x8 and 8  e =  ln 8    x ln 8 x ln 8 1 1  1 1 e , then g( x )  x e . Thus, g      e   .  ln 8  ln 8 eln 8Sign of g( x ) :  +  2  ln 8 2  The function g is concave upward on the interval   ,   and is  ln 8   2  concave downward on the interval   ,   .  ln 8 2 The function g has an inflection point at x   and the y-coordinate ln 8  2  2 2 g      e  2   x of the point is   2 using that g( x )  x8  ln 8  ln 8 e ln 8 = x e x ln 8 .Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860  1   1  Answer: Increasing:   ,   Decreasing:   ,    ln 8   ln 8 1 Local Maximum: None Local Minimum:  eln 8 2   2  CU:   ,   CD:   ,    ln 8   ln 8  2 2    ,   Inflection Point:  2   ln 8 e ln 8 Example Find the equation (in point-slope form) of the tangent line to the graph of 3 the function y  ( 2 x 2  15)e x  4 x  6 at the point for which x  2 .Tangent Point: ( 2, y( 2))  ( 2,  7e 6 ) y( 2)  (8  15)e 8  8  6   7e 63 3 y  4 x e x  4 x  6  ( 2 x 2  15)e x  4 x  6 (3 x 2  4)6 6 6 6 6 m tan  y( 2)  8e  (  7)e (8)  8e  56e   48eAnswer: y  7e 6   48e 6 ( x  2)Example Find the equation (in point-slope form) of the normal line to the graph of the function y  x 2 4 x at the point for which x   3. 9  Normal Point: (  3, y(  3))    3,   64  1 9 y(  3)  9( 4  3 )  9   64 64Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 y  2 x 4 x  x 2 4 x ln 4 = x 4 x ( 2  x ln 4)3( 2  3ln 4) m  y(  3)   3(4  3 )( 2  3ln 4) =  tan 641 64 m normal   = m tan 3(2  3ln 4)9 64 9 64 Answer: y   ( x  3) or y   ( x  3) 64 3( 2  3ln 4) 64 3( 2  ln 64)D e x e x e x dx  e x  c Since x  , then  . b x Since D b x  b x ln b , then b x dx   c . x  ln bExamples Evaluate the following integrals.2 1.  x e5 x dxLet u  5 x 2 Then du  10 x dx2 1 2 1 1 1 2 x e5 x dx = 10 x e5 x dx = eu du = eu  c = e5 x  c  10  10  10 101 2 Answer: e5 x  c 1015e 3t dt 2.  3t 3 27  8eCopyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Let u  27  8e 3t Then du   8e 3t 3dt   24e 3t dt15e 3t e 3t 15  24e 3t dt 15 dt dt  3t =  3t =  3t = 3 27  8e 3 27  8e  24 3 27  8e5 du 5 5 3 15   =  u  1/ 3 du =   u 2 / 3  c =  u 2 / 3  c = 8 3 u 8  8 2 1615  ( 27  8e 3t )2 / 3  c 1615 Answer:  ( 27  8e 3t )2 / 3  c 164 3. e 3 x  2 dx   1The integrand f ( x )  e 3 x  2 is continuous on the closed interval [  1, 4] . Thus, we can apply the Fundamental Theorem of Calculus to this integral.Let u  3 x  2 Then du  3dx10 4 3 x  2 1 4 1 10 1  e dx = e 3 x  2 3dx = e u du  e u =   1   1   5  3 3 3   51 1 e 15  1 ( e 10  e  5 ) = e  5 ( e 15  1) = 3 3 3e 5NOTE: Since u  3x  2 , then when x   1, we have that u   5. Thus, the new lower limit of integration is  5 . Similarly, when x  4 , we have that u  10 . Thus, the new upper limit of integration is 10.Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 e 15  1 Answer: 3e 5 t 2 4. 3 dt  5 t  82 t 2 8  t 3 3 dt = t 5 dt  5 t  8 Let u  8  t 3 Then du   3t 2dt3 1 3 1 t 2 5 8  t dt =  (  3t 2 )5 8  t dt =  5 u du =  3  3 3 1 5 u 5 8  t    c =   c 3 ln 5 3ln 53 3 5 8  t 5 8  t Answer:   c or   c 3ln 5 ln 125 w 9 e 5. dw  5 w e w The integrand f ( w)  is continuous on the closed interval [5, 9]. w Thus, we can apply the Fundamental Theorem of Calculus to this integral. Note that the integrand does have a discontinuity at w  0 .1 NOTE: If you know that D x x  , then you would try the 2 x following substitution.Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Let u  w 1 Then du  dw 2 w w w 9 e 9 e 3 3 dw = 2 dw = 2 e u du  2e u  =  5 w  5 2 w  5 52( e 3  e 5 )NOTE: Since u  w , then when w  5 , we have that u  5 . Thus, the new lower limit of integration is 5 . Similarly, when w  9 , we have that u  9  3 . Thus, the new upper limit of integration is 3.Answer: 2( e 3  e 5 )2 x 6. dx  7 x1 NOTE: If you know that D x x  , then you would try the 2 x following substitution.Let u  x 1 Then du  dx 2 x2 x 1 2 x 1 2 x 2 dx = dx = 2  dx = 2 u du =  7 x 7  x 7  2 x 7 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 2 2 u 2 2 x 2  2 x 2 1  x   c =   c =  c =  c 7 ln 2 7 ln 2 7 ln 2 7 ln 22 1  x Answer:  c 7 ln 27.  3 x sin 3 x cos 3 x dxThere are three ways to work this integral. a. Let u  sin 3 x x x x x Then du  (cos 3 )( D x 3 )dx = (cos 3 )(3 ln 3)dx1 x x x 3 x sin 3 x cos 3 x dx = 3 (ln 3)sin 3 cos 3 dx =  ln 3 2 1 1 u 1 1 2 x u du =   c = sin 2 3 x  c = sin 3  c ln 3  ln 3 2 2 ln 3 ln 91 2 x Answer: sin 3  c ln 9 b. Let u  cos 3 x x x x x Then du  (  sin 3 )( D x 3 )dx = (  sin 3 )(3 ln 3)dx1 x x x 3 x sin 3 x cos 3 x dx =  3 (ln 3)(  sin 3 )cos 3 dx =  ln 3 2 1 1 u 1 2 x  u du =    c =  cos 3  c = ln 3  ln 3 2 2 ln 3Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 1  cos 2 3 x  c ln 91 2 x Answer:  cos 3  c ln 9 NOTE: In order to show that this answer is the same as the answer above, we will use the Pythagorean identity cos 2   sin 2   1 for all  . Thus, cos 2 3 x  sin 2 3 x  1 for all x. Thus, cos 2 3 x =1 2 x 1  sin 2 3 x for all x. Thus,  cos 3  c = ln 9 1 1 1  (1  sin 2 3 x )  c =   sin 2 3 x  c = ln 9 ln 9 ln 9 1 1 1 1 sin 2 3 x  c  = sin 2 3 x  k , where k  c  . ln 9 ln 9 ln 9 ln 9 c. By the double angle formula for sine, we have sin 2 = 1 2sin  cos  . Thus, sin 3 x cos 3 x  sin ( 2  3 x ) . Then 2 1  3 x sin 3 x cos 3 x dx = 3 x sin ( 2  3 x ) dx =    2 1 3 x sin ( 2  3 x ) dx 2 Let u  2  3 x Then du  2(3 x ln 3)dx = 3 x ln 9dx1 1 1 x x 3 x sin ( 2  3 x ) dx =  3 (ln 9)sin ( 2  3 ) dx = 2  ln 9 2 1 1 1 sin u du = (  cos u )  c =  cos ( 2  3 x )  c 2 ln 9  2 ln 9 2 ln 9Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 1 Answer:  cos (2  3 x )  c 2 ln 9NOTE: Recall the double angle formula for cosine is given by cos 2 = cos 2   sin 2  . If you replace cos 2  by 1  sin 2  , you obtained that cos 2 = 1  2sin 2  . If you use this form of the double angle formula for cosine, then you can verify that the answer in (c) is the same as the answer in (a). If you replace sin 2  by 1  cos 2  , you obtained that cos 2 = 2cos 2   1. If you use this form of the double angle formula for cosine, then you can verify that the answer in (c) is the same as the answer in (b). 8. 6 cot ( / 4) csc 2 d  4 Let u  cot 4 1  Then du   csc2 d 4 4  1   u 6 cot ( / 4) csc 2 d =  4 6 cot 4   csc 2  d =  4 6 du =  4   4 4  6 u 4  6 cot ( / 4)  4   c =   c ln 6 ln 64  6 cot ( / 4) Answer:   c ln 6Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860

Lesson 2 Derivatives and Integrals Involving The

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