Problems about Limits and Differentiation of functions

• READ THIS FIRST
  
• Problems about Differentiation of functions and limits
  
  • Level 1 problems
    
  • Level 2 problems
    
  • Level 3 problems
    
• External links to solved problems on differentiation
  
  • Chain Rule
    
  • Product rule
    
  • Quotient rule
    
  • Trigonometric functions
    
  • Cyclometric functions
    
  • Implicit differentiation
    
  • Functions u^v
    

READ THIS FIRST

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Differentiation of functions and limits

Level 1 problems

 
y = 5x  ;  y' = 5

y = x3  ;  y' = 3x2

y = (2x2+ x + 5)  ;  y'= 4x + 1

     x - 1            (x+1).1 - (x-1).1             2
y = ------- ;   y' = -------------------- = ------------------
     x + 1              (x+1)(x+1)              (x+1)(x+1)

y = (x+5)6  ;  y' = 6.(x+5)5


y = (2x + 6)3 ; y' = 3.(2x + 6)2.2

y =  (2x + 6)(5x - 7) ; y'= 2.(5x - 7) + (2x + 6).5

y =  (2x + 6)2.(5x - 7)3; y'= 2.(2x + 6).2.(5x - 7)3+ (2x + 6)2.3.(5x - 7)2.5

y = x/5  ; y' = 1/5

                    1
y = 5/x  ; y' = -5.---
                   x.x

     7
y = ---  <=> y = 7.x-4  ;   y' = -28 x-5
    x4

Level 2 problems

• 
  

  Calculate the derivative of ax + b y = --------- cx + d

  
  

   
               (cx + d)a - (ax + b)c       ad - bc
          y' = ---------------------  = ----------------
                  (cx + d)(cx + d)      (cx + d)(cx + d)
  

• 
  

  Calculate the derivative of ad - bc y = ---------------- (cx + d)(cx + d)

  
  

   
  Since (ad -bc) is constant we write y as
  
          y = (ad -bc). (cx + d)-2
  
  Then
          y' = (ad -bc).(-2) (cx + d) -3 .c
  
  So,
                  -2c(ad -bc)
          y' = -----------------
                  (cx + d)3
  

• 
  

  Calculate the first and second derivative of x.x + x + 1 y = ------------- x + 1

  
  

   
  
      (2x+1)(x+1)-(xx+x+1)      x(x+2)
  y' = -------------------- = ------------
       (x+1)(x+1)              (x+1)(x+1)
  
       (2x+2)(x+1)2-2(x+1)(x2+2x)         2
  y" = --------------------------- = ---------
                  (x+1)4              (x+1)3
  

• 
  

  Calculate the first and second derivative of y = sqrt(x3-3x +2)

  
  

   
                  3x2-3               3(x2-1)
          y' = ----------------- =  ----------------------
               2.sqrt(x3-3x +2)     2.sqrt( (x-1)2(x-2) )
  
  
  It is difficult to  calculate the second derivative from this result.
  Therefore we simplify the result in the following way.
  For all x > 1
                 3    (x +1)
          y' =  ----.-----------
                 2    sqrt(x-2)
  Then
                3   sqrt(x-2) -(x +1).1/(2.sqrt(x-2) )
          y" = ---. ---------------------------------- = ...
                2       (x-2)
  
                 3   (x + 3)
          y" =  ---.---------------
                 4   (x-2)sqrt(x-2)
  For all x < 1
                  3    (x +1)
          y' = - ---.-----------
                  2    sqrt(x-2)
  Then
                  3   sqrt(x-2) -(x +1).1/(2.sqrt(x-2) )
          y" = - ---. ---------------------------------- = ...
                  2       (x-2)
  
                  3   (x + 3)
          y" = - ---.----------------
                  4   (x-2)sqrt(x-2)
  
  

• 
  

  Calculate the first and second derivative of y = sin(2x) - cos(2x)

  
  

   
          y' = 2cos(2x) + 2sin(2x)
          y" = -4sin(2x) + 4cos(2x)
  

• 
  

  An area A depends on x in the following way. 2r3.x3 A = -------------- with r = constant. (x2 + r2)2 Calculate A'.

  
  

   
                      (x2 + r2)2.3x2- x3.2(x2 + r2).2x
          A' = 2r3 . -------------------------------------
                          (x2 + r2)4
  
  
             =  ...
  
                      x2.(3r2 - t2)
          A' =  2r3.-----------------
                       (x2 + r2)3
  

• 
  

  Find 1 - sqrt(cos(x)) lim ------------------ 0 x2

  
  

   
  With l'Hospital's rule
  
                sin(x)/(2.sqrt(cos(x)))
          = lim ------------------------
             0            2x
  
                 sin(x)       1
          = lim -------.------------
             0     x    4sqrt(cos(x))
  
  
          = 1.(1/4) = 1/4
  

• 
  

  Find x(2x + 3)x - 1 lim ------------------- -1 x3 - 3x - 2

  
  

   
  
  First method:
                 (x+1)(x+1)(2x-1)
          = lim ---------------
            -1   (x+1)(x+1)(x-2)
  
                 (2x-1)
          = lim -------- = 1
            -1   (x-2)
  
  Second method:
               2x3 + 3x2 -1
          lim -------------------
          -1   x3 - 3x - 2
  
                  With l'Hospital's rule
  
                  6x2 + 6x
          = lim -------------
            -1    3x2 - 3
                  With l'Hospital's rule
  
                  12x + 6
          = lim -------------  = 1
            -1      6x
  

• 
  

  sin(3x) + tan(2x) Find lim ------------------ 0 x - 2 sin(x)

  
  

   
  
                sin(3x) + tan(2x)
            lim ------------------
            0     x - 2 sin(x)
  
  
                sin(3x)      tan(2x)
               -------- .3 + ------  .2
                 3x           2x              5
        =  lim ------------------------ =  ------- = -5
            0              sin(x)            -1
                  1  - 2 ---------
                             x
  

• 
  

  x3 ex + 2x2 - 3 Find lim -------------------- + infty x2 - e2x + 5

  
  

   
   Use l'Hospital's rule  3 times
   Then you have after simplification
  
                    6 + 18x + 9x2 + x3
          lim     ----------------------
        + infty          -8 ex
  
   Use l'Hospital's rule again  3 times
   Then you have after simplification
                6
          lim ------- = 0
       +infty -8 ex
  

  The numbers a and x are positive. Find : x sqrt(x) - a sqrt(a) lim ------------------------- a sqrt(x) - sqrt(a)

  
  

  Case 0/0. First we write the limit in a modified form, then we apply l'Hospital's rule.

   
           x3/2 - a3/2
     lim ----------------------
      a    x1/2 -  a1/2
  
  
          3/2 x1/2
   = lim ------------
      a   1/2 x-1/2
  
  
  
   = 3 a
  
  

Level 3 problems

• 
  

  x2 + 2px + q Given : y = -------------- x2 + 1 Prove that there are two x values, x' and x", such that y' = 0. Then prove that x'.x" = -1.

  
  

   
               (x2 + 1)(2x + 2p) - (x2 + 2px + q).2x
          y' = --------------------------------------
                  (x2 + 1)2
  
  
             = ...
  
                  -2px2 + 2(1 - q)x +2p
             = -------------------------
                  (x2 + 1)2
  
          y' = 0 if and only if   -2px2 + 2(1 - q)x +2p = 0
  
          This gives two values x' and x".
  
          It is immediate that x'.x" = -1
  

• 
  

  The derivative of f(x) is f'(x). The derivative of f'(x) is f"(x) an is called the second derivative of f(x). The derivative of f"(x) is f"'(x) an is called the third derivative of f(x). ... Now, let f(x) = sqrt(x) n-1 (2n-2)! (1-2n)/2 Prove that the n-th derivative of f(x) = (-1) .--------.(4x) (n-1)!

  
  

   
  f(x) = x1/2
  
  f'(x) = (1/2). x(-1/2)
  
  f"(x) =(1/2)(-1/2) . x(-1/2-1)
  
  f"'(x) =(1/2)(-1/2)(-1/2 - 1) . x(-1/2-2)
  
  f""(x) =(1/2)(-1/2)(-1/2 - 1)(-1/2 - 2) x(-1/2-3)
  
  ...
  
  So, the n-th derivative of f(x)
  =(1/2)(-1/2)(-1/2 - 1)(-1/2 - 2)...(-1/2 -(n-2)) x(-1/2-(n-1))
  
        n-1   1
  = (-1)   .----.(1+2).(1+2.2)(1+3.2)...(1+(n-2).2). x(-1/2-(n-1))
             2n
  
        n-1   1
  = (-1)   .----.3.5.7.9...(2n-3). x(1-2n)/2
             2n
  
        n-1   1   1.2.3.4.5.6....(2n-3)(2n-2)
  = (-1)   .----.---------------------------- x(1-2n)/2
            2n   1.2.4.6. ...  (2n-2)
  
  
  
        n-1   1   1.2.3.4.5.6....(2n-3)(2n-2)
  = (-1)   .----.---------------------------- x(1-2n)/2
             2n    2n-1. 1.2.3... (n-1)
  
  
        n-1     (2n-2)!
  = (-1)   .-------------------- x(1-2n)/2
             22n-1   .(n-1)!
  
  
        n-1     (2n-2)!
  = (-1)   .------------.2 (1-2n) x(1-2n)/2
                (n-1)!
  
  
        n-1     (2n-2)!
  = (-1)   .------------.(4x)(1-2n)/2
                (n-1)!
  
  

• 
  

  Find lim (x. sin(3/x)) x -> infty

  
  

   
                  sin(3/x)
          = lim -----------
      x -> infty  (1/x)
  
                  Let t = 1/x
  
                  sin(3t)
          = lim -----------
             0      t
  
                3.cos(3t)
          = lim ----------- = 3
             0      1
  

• 
  

  The function f(x) = ( (x-2)2 )1/3 - 3 is given in interval [1,3]. Investigate whether or not we can apply Rolle's theorem.

  
  

  First condition: f(a)=f(b)?

  Since f(1)= -2 and f(3) = -2, the first condition is satisfied.

  Second condition: continuous in [a,b] ?

  f(x) is an algebraic function. Therefore is is continuous in its domain. Thus f(x) is continuous in [1,3].

  Third condition: differentiable in ]a,b[ ?

  • x > 2

     
       y = (x-2)2/3 - 3
    
                 2
       y' = ---------------- > 0
             3 (x-2)1/3
    
       lim  y' = + infty
      x -> 2
    

  • x < 2

     
       y = (2-x)2/3 - 3
    
                - 2
       y' = ---------------- < 0
             3 (2-x)1/3
    
       lim  y' = - infty
      x -> 2
    

    The function is not differentiable in 2. The condition is not satisfied. On a graph the result is immediately obvious.
    
• 
  

  The function ex.sin(x) satisfies the three conditions of Rolle's theorem in [0,pi]. Find point P(c,f(c)) such that f'(c) = 0.

  
  

   
      y' = ex.sin(x) + ex.cos(x)
  
      y' = 0
  <=>
      sin(x) + cos(x) = 0
  <=>
      sin(x) = - cos(x)
  <=>
      tan(x) = -1
  
  In  [0,pi] the solution is   x = 3 pi/4. So, point P = P(2.36 , 7.46).
  

External links to solved problems on differentiation

Chain Rule

Product rule

Quotient rule

Trigonometric functions

Cyclometric functions

Implicit differentiation

Functions u^v