Calculus Mid-Term Review          Focus: Chapter 3 Derivatives of a Function 
          
         Concepts:  Differentiability (do left- and right-hand derivatives equal?) 
                     4 (actually 3) ways a function might fail to have a derivative 
                     Intermediate Value Theorem for Derivatives 
                     Rules for Differentiation (“shortcuts”) 
                     Power/Product/Quotient/Chain Rules 
                     Second and Higher Order Derivatives 
                     Position/Velocity/Acceleration Hierarchy and definition of Speed 
                     Derivatives of Trigonometric Functions 
                     Implicit Differentiation 
                     Derivatives of Exponential and Logarithmic Functions 
                      
         Problems: 
         Find all the values of x for which the function is differentiable. 
                     3x2 −3x+5
         1)  f (x) =     x −4       all reals except x = 4                                  
                
         2)  f (x) =  x −5    (5, ∞)               
                
 €       3)  f (x) = 3x −5    all reals except x = 5/3 
                
 €        
 €        
         Find dy/dx. 
                   2                                  x−4 6x−3 − 3x2−3x+5 1
         4)  y = 3x −3x +5                      dy = (     )(       )  (             )( )  
                     x −4                      dx                  (x−4)2
          
         5)  y = 3x2cos 3x2                     dy =3x2 −sin(3x2) (6x)+6xcos(3x2) 
 €                      (    )                 dx        (           )
          
         6)  y = 3ex2x2                         dy =3ex2 (2x)+3ex2 (2x)(x2) 
 €                                             dx
          
         7)  y = 3x5 + 2x4 + 8x3 −5x +4         dy =15x4+8x3+24x2−5 
 €                                             dx
          
         8)  y =  x2 −4x+1                      dy = 1 x2 −4x+1 −1/2(2x−4) 
 €                                             dx    2(            )
          
                 3tan(x2)                       dy   2x(3sec2(x2))(2x)−3tan(x2)(2)
 €       9)  y =    2x                         dx =                  4x2                    
                
          
          
 €        
       10)  y = ln(cos(3x4))            dy =    1     −sin 3x4   12x3 =−12x3tan 3x4  
                                        dx   cos(3x4)(     (   ))(    )           (   )
                                        dy      1            3
  €    11)  y = log 3x3                    =        (9x2)=        
                  4(   )                dx   3x3ln4         xln4
        
       12)  y = 2cosx5                  dy =2cosx5 ln2(−sinx5)(5x4) 
  €                                     dx
        
       13) Find all of the nonzero derivatives of  y = 3x5 + 2x3 + 5x −2 
  €            4    2                             
         !
        y =15x +6x +5
         !!     3
        y =60x +12x
         !!!     2                   € 
        y =180x +12       
        y(4) = 360x
        y(5) = 360
        
       Use implicit differentiation to find dy/dx: 
       14)  x4 − y4 = 3x + 2y                      15)  xy + 3x −2y =5 
                                                           
        4x3−4y3 dy =3+2dy                               x dy + y+3−2dy =0
  €              dx       dx                 €           dx          dx
        4x3−3=2dy+4y3dy                                 x dy −2dy =−y−3  
                  dx      dx                             dx    dx
        dy   4x3−3                                      dy  −y−3
        dx = 2+4y3                                      dx = x−2
        
        
       16) cosx+sin y = x2y3 
               
        −sinx+cosy"dy%=x2"3y2 dy%+2xy3
                    $   '    $      '
                    #dx&     #    dx&
  €     cosy"dy%−3x2y2 dy =2xy3+sinx        
            $   '
            #dx&         dx
        dy   2xy3+sinx
        dx = cosy−3x2y2
        
        
                                    2   2
       Use implicit differentiation to find d y/dx : 
       17)  x2 = 2− 3                18)  x2 − x = 3sin y 
                 y                          
               3 dy                      2x−1=3cosydy
        2x=0+                                        dx
               y2 dx            €          dy   2x−1
  €     dy  2                              dx = 3cosy
        dx = 3 xy2
                                                3cosy 2 − 2x−1 −3siny "dy%
       d2y  2 !   dy$ 2                   d2y  (     )( ) (    )(      )$  '
          = x#2y    &+ y2                     =                         #dx&    
       dx2  3 "   dx% 3                   dx2              9cos2 y
          = 4 xy!2 xy2$+ 2 y2                                           " 2x−1%
                #    &                          3cosy 2 − 2x−1 −3siny
            3   "3   % 3                       (     )( ) (    )(      )$3cosy'
                                              =                         #     &
          = 8 x2y3 = 2 y2                                   9cos2 y
            9      3                            6cosy+ 2x−1 2tany
                                              =       (     )
                                                     9cos2 y
       Calculus Mid-Term Review          Focus: Chapter 2 Limits and Continuity 
        
       Concepts:  Definition of a Limit 
                  Properties of Limits 
                  The Sandwich Theorem 
                  Horizontal Asymptotes 
                  Vertical Asymptotes 
                  End Behavior Model 
                  Continuity at a point 
                  Intermediate Value Theorem for Continuous Functions 
                  Tangent and Normal Lines to a curve 
                   
       Problems: 
              x2 − x+1  1−1+1 1
       1) lim          =       =                          
           x→1  x+3       1+3    4
              x−1            x−1           1    1
       2) lim 2    =lim              =lim      =          
           x→1 x −1   x→1 x−1 x+1      x→1 x+1  2
                         (    )(   )
        
              (2+x)3−8        8+12x+6x2+x3−8
       3) lim           =lim                     =lim 12+6x+x2 =12            
           x→0    x       x→0         x            x→0 (          )
        
          lim x2 −5 = lim#x+1+ −4 &=∞
       4)  x→1        x→1%          (                     
               x−1       $      x−1'
        
        
              x2 +3x−4      x−1 x+4
       5) lim           lim(    )(    ) = lim(x+4)=5   
           x→1   x−1    x→1    x−1        x→1
        
          lim3sin2 x−4sinxlimsinx(3sinx−4) =lim 3sinx−4 =−4
       6)  x→0     sinx     x→0      sinx        x→0 (       )      
        
        
       7) Use the diagram to the right to answer the following: 
        
       A)  lim f (x) = 1 
           x→2−
              
        
       B)  lim f (x) = -2 
 €         x→2+
              
        
       C) lim f (x) = DNE 
 €         x→2
              
        
 €     D) f (2) = -2