Differentiation of functions,limits (II), maximum, minimum, inflection points.




Two special limits of trigonometric functions

Without proof, we accept that

 
            sin(x)
        lim ------- = 1
      x->0    x

 
            tan(x)         sin(x)         sin(x)    1
        lim ------- = lim --------- = lim ------ .------- = 1.1 = 1
         0    x        0   x.cos(x)         x      cos(x)
 
            tan(x)
        lim ------- = 1
      x->0    x

Related limits of trigonometric functions

Many related limits can be calculated relying on previous formulas
 
         sin(5x)
   lim ----------- = 1
    0       5x

         tan(53x)
   lim ----------- = 1
    0       53x


           5x
   lim ----------- = 1
    0    sin(5x)

           3x             (3/5) (5x)
   lim ----------- = lim --------------- = 3/5
    0    sin(5x)      0     sin(5x)

           5 x2          (1/5) (5x)  (5x)
   lim ----------- = lim  ------------------- = 1/5
    0    sin2(5x)    0      sin(5x) sin(5x)


         sin(5x)            sin(5x) / (5x)
   lim ----------- = lim ------------------ = 1
    0    tan(5x)      0     tan(5x) / (5x)

         sin(3x)           3 sin(3x) / (3x)
   lim ----------- = lim ---------------------    = 3/2
    0    tan(2x)      0    2 tan(2x) / (2x)


         tan2(4x)        16 (tan(4x)/(4x)) (tan(4x)/(4x))
   lim ----------- = lim -------------------------------- = 16
    0    x sin(x)                   sin(x)/x


   lim ( cot(x) - 1/x) = lim ( 1/tan(x) - 1/x)
    0                     0

           1 - tan(x)/x      1 - 1
   = lim (--------------) = ------- = 0
      0      tan(x)/x          1
All these limits can also be calculated using de l'Hospital rule. See below.

The derivative of

f(x) for x = a

Say G is the graph of f(x) and point P(a,f(a)) is on G. Now, take a point Q(a+h,f(a+h)) on G close to point P.
 
                    (f(a+h)-f(a))
The slope of PQ is ---------------
                         h

This slope can be viewed as the mean slope of G in the curve segment [PQ].
Now, let h become smaller and smaller. Then the line PQ approaches to the tangent line to the curve at point P.
If the following limit exists and if it is real, then

 
                                      (f(a+h)-f(a))
the slope of that tangent line = lim --------------
                                h->0        h
The last limit is called the derivative of f(x) at point P. This value is noted f'(a).

f(x)

Now make the point P from previous section variable. Then
 
        the slope of the tangent line in point P(x,f(x))

               (f(x+h)-f(x))
        = lim ---------------
         h->0        h

        = the derivative of f(x)

        = f'(x)        (notation)


        = D(f(x))       (notation)

           df
        = ---           (notation)
           dx

           d
        = --- f(x)      (notation)
           dx
If, for x = b, this limit does not exist or if it is not a real number, then we say that the derivative does not exist for x = b, or that f(x) is not differentiable for x = b.

A function is differentiable in [a,b] if it is differentiable for each x in [a,b].

Differentiable and continuous

Theorem:
If a function f(x) is differentiable for x = b, then f(x) is continuous for x = b.

Proof:
Since f(x) is differentiable for x = b, we have

 

              (f(b+h)-f(b))
         lim --------------- = a real number = f'(b)
         h->0        h
Now,


                          f(b+h) - f(b)
      lim f(b+h) = lim ( ---------------- .h + f(b)  )
      h->0        h->0          h

                           f(b+h) - f(b)
                    lim ( ---------------- .h )  + f(b)
                 = h->0          h


                 = f'(b) .0  + f(b)

                 = f(b)

Let x = b+h ; if h ->0  then x -> b

      lim f(x) = f(b)
     x->b

And f(x) is continuous in b.

Formulas to calculate the derivative of

A constant function

Take f(x) = c.
The slope of the tangent line in point p of a constant function is 0. So,
 
        d
        -- c = 0    or    D(c) = 0
        dx

Examples: D 3 = 0 ; D sin(4) = 0

f(x) = x

The slope of the tangent line in point p of the function x is 1. So,
 
        d
        -- x = 1   or D(x) = 1
        dx

The sum f(x) + g(x)

If the functions f(x) and g(x) are differentiable then
 
d                     f(x+h)+g(x+h) -(f(x)+g(x))
-- (f(x)+g(x)) = lim -----------------------------
dx              h->0             h

                      (f(x+h)-f(x)) + (g(x+h)-g(x))
                = lim -----------------------------
                 h->0              h

                        (f(x+h)-f(x))            (g(x+h)-g(x))
                =  lim ---------------  +   lim ---------------
                   h->0        h            h->0        h

                  d         d
                = -- f(x) + -- g(x)
                  dx        dx
 
d                d         d
-- (f(x)+g(x)) = -- f(x) + -- g(x) = f'(x) + g'(x)
dx               dx        dx

Example: D(x + 1/pi) = 1 + 0 = 1

This property is extendable to the sum of n differentiable functions.

The product f(x).g(x)

If the functions f(x) and g(x) are differentiable then
 
d                     f(x+h).g(x+h) - f(x).g(x)
-- (f(x).g(x)) = lim ---------------------------
dx              h->0             h

                       f(x+h).g(x+h) -f(x+h)g(x) + f(x+h)g(x) - f(x).g(x)
                = lim --------------------------------------------------
                 h->0                     h

                                 (g(x+h)-g(x))          (f(x+h)-f(x))
                =  lim (f(x+h). --------------- + g(x). --------------)
                   h->0               h                       h

                = f(x).g'(x) + f'(x).g(x)
 
d
-- (f(x).g(x)) = f(x).g'(x) + f'(x).g(x)
dx

The product of n differentiable functions.

Previous property is extendable to the product of n differentiable functions.
Example : u=f(x) v=g(x) w=h(x)

 
d
--(u.v.w) = u'.v.w + u.v'.w + u.v.w'
dx


Special product

u=f(x) and c is a constant

 
d
--(c.u) = c.u'
dx

Examples:

 
  D(3x) = 3

  D(2x + 13) = 2

  D(x2/434) = 1/434 D(x2)

Power of f(x)

u=f(x) is differentiable
 
d    n    d                       n-1
--( u ) = --(u.u.u. ... .u) = n.(u   ).u'
dx        dx
 
d
--( un ) = n.(un-1).u'
dx

Examples:

 
  D(x3) = 3 x2

  D(x8) = 8 x7

  D(x3 + x4) = 3x2 + 4x3

  D(x3 + 34) = 3 x2

  D( (12x)3 ) = 3(12x)2.(12)

  D( (12x + 4)4 ) = 4(12x +4)3.12

  D( (12x+4)(x4+5)) = D(12x+4) .(x4+5) + (12x+4).D(x4+5) = 12.(x4+5) + (12x+4).4x3

  D( (x3.(x+1)5) = 3x2.(x+1)5 + x3.5(x+1)4

  D( (x3.(x2+1)5) = 3x2.(x2+1)5 + x3.5(x2+1)4.2x

The quotient f(x)/g(x)

If the functions f(x) and g(x) are differentiable then
 
d                     f(x+h)/g(x+h) -(f(x)/g(x))
-- (f(x)/g(x)) = lim -----------------------------
dx              h->0             h

                        f(x+h).g(x) - f(x).g(x+h)
                =  lim -----------------------------
                  h->0      h.g(x).g(x+h)

                        f(x+h).g(x) -f(x)g(x) +f(x)g(x) - f(x).g(x+h)
                =  lim ---------------------------------------------
                  h->0               h.g(x).g(x+h)


                       g(x) . (f(x+h)-f(x))/h - f(x) . (g(x+h)-g(x))/h
                =  lim ------------------------------------------------
                  h->0             g(x).g(x+h)

                    g(x).f'(x) - f(x).g'(x)
                =  -------------------------
                             g(x).g(x)
 
      (u / v)' = ( v u' - u v') / v2

Example :

 
    d   2x+7      (3x+2)2 - (2x+7)3
   -- (------) = -----------------------
   dx   3x+2         (3x+2)2

Important special cases

u=f(x) is differentiable
 
                        n         (n-1)
d    -n    d    1      u . 0 - n.u     .u'        (-n-1)
--( u  ) = -- (---) = --------------------- = -n.u      .u'
dx         dx    n               2n
                u              u

From this it follows that the formula
 
d    n         n-1
--( u ) =  n.(u   ).u'
dx
holds for all integer values of n.

Here are some additional useful formulas:

 
  D( 1/x) = D( x-1 ) = -1.x-2 = -1/x2

 d          d
--(1/u ) = --(u-1 ) = -1.u-2.u'
dx         dx

             -1
         = ----- u'
            u2
 
 d          -1
--(1/u ) = ----- u'
dx          u2

Examples:

 
  D( 1/(2x+23) ) = -2/(x+23)2

        1         -1
  D( -------) = ---------- .4. 5x4
       4x5      (4x5)2

        1           -1
  D ---------- = ---------- . 4(2x+23)3. 2
     (2x+23)4    (2x+23)8


  D (x2+4x)3 = 3(x2+4x)2 (2x+4)


  D ((x-1)(x+1)(x2+1)) = D ((x2-1)(x2+1)) =D (x4-1) = 4x3


  D (1/(2x+4)3) = D (2x+4)-3 = -3 (2x+4)-4.2 = -6/(2x+4)4

sin(x)

 
d              sin(x+h) - sin(x)
--sin(x) = lim ------------------
dx        h->0         h

               2.cos(x + h/2).sin(h/2)
         = lim ------------------------
          h->0         h

               cos(x + h/2).sin(h/2)
         = lim ------------------------
          h->0        h/2

                            sin(h/2)
         = lim cos(x + h/2).--------
          h->0                h/2


         = cos(x)

cos(x)

Analogous as for sin(x) you can prove that
 
d
--cos(x) = - sin(x)
dx

tan(x)

 
d          d  sin(x)    cos2(x) - sin(x).(-sin(x))
--tan(x) = --(------) = -----------------------------
dx         dx cos(x)        cos2(x)

             1
        = ---------
          cos2(x)

cot(x)

Analogous as for tan(x) you can prove that
 
d              -1
--cot(x) =  ---------
dx           sin2(x)

The chain rule

Take the function f(g(x)) en let u = g(x). Assume that f and g are differentiable.
 
d                f(g(x+h)) - f(g(x))
--(f(g(x)) = lim ------------------
dx          h->0        h

             f(g(x+h)) - f(g(x))    g(x+h) - g(x)
      =  lim --------------------. ----------------
        h->0   g(x+h) - g(x)              h

                    let g(x+h) = g(x) + k = u + k

              f(u + k) - f(u)      g(x+h) - g(x)
      =  lim -------------------. ----------------
        h->0       k                     h

                    if h -> 0 then k -> 0

              f(u + k) - f(u)           g(x+h) - g(x)
      =  lim -------------------.  lim ----------------
        k->0       k              h->0        h

         d         d
      =  -- f(u) . -- g(x)
         du        dx
We can also write

 
d         d
-- f(u) = -- f(u) . u'
dx        du
This formula is known as the chain rule.

Corollaries

Appealing on the chain rule we have
 
d          d
--sin(u) = -- sin(u) . u' = cos(u) . u'
dx         du

d          d
--cos(u) = -- cos(u) . u' = - sin(u) . u'
dx         du

d          d                    1
--tan(u) = -- tan(u) . u' =  ---------- . u'
dx         du                 cos2(u)

d          d                    1
--cot(u) = -- cot(u) . u' = - -------- . u'
dx         du                 sin2(u)

d
--(un ) = n . (un-1).u'
dx

...
 
d
--sin(u) =  cos(u) . u'
dx

d
--cos(u) = - sin(u) . u'
dx

d              1
--tan(u) =  ---------- . u'
dx           cos2(u)

d              1
--cot(u) = - -------- . u'
dx           sin2(u)

d
--(un ) = n . (un-1).u'
dx


Examples:

 
 D( sin(x2) ) = cos(x2) . 2x


                         3
 D( tan(3x+2) ) = ---------------
                   cos2(3x+2)


 D ( x2 . sin(3x) ) = 2x . sin(3x) + x2. cos(3x).3

 D ( sin(x)/x) = D( (1/x).sin(x)) = (-1/x2).sin(x) +(1/x).cos(x)

                          1
 D tan(sqrt(x3)) = ---------------. (3/2) x1/2
                     cos2(x3/2)

                         1
 D tan(sin(1/x)) = ------------------ cos(1/x) . (-1/x2)
                    cos2(sin(1/x))

 D sin3(pi/2 - x) = D cos3(x) = 3cos2(x) (-sin(x))

      1
 D -------- = D cos-2(x) = -2 cos-3(x).(-sin(x))
   cos2(x)


 D sinm(xm) = m sinm-1(xm) cos(xm) m xm-1


        x2 - 1         x2 - 1      (x2 + 1)2x -(x2 - 1)2x
 D sin(--------) = cos(--------) . -------------------------- = ...
        x2 + 1         x2 + 1            (x2 + 1)2

The inverse trigonometric functions

First take arcsin(x) . We know that for x in [-1,1] sin( arcsin(x) ) = x. (1)
So, these two functions must have the same derivative.
We differentiate both sides of (1).
From previous formulas we have
 
                       d
     cos( arcsin(x) ). --( arcsin(x) ) = 1      (2)
                       dx
But let b = arcsin(x). Then b is in [-pi/2,+pi/2] .
 
            _____________      _______
           |        2         |      2
 cos(b) = \| 1 - sin (b)  =  \| 1 - x
                            ________
                           |      2
Thus, cos( arcsin(x) ) =  \| 1 - x
From (2) we can write now :
 
          ________
         |      2     d
        \| 1 - x    . --( arcsin(x) ) = 1
                      dx

       d                     1
<=>    --( arcsin(x) ) = -------------
       dx                   _______
                           |      2
                          \| 1 - x

and with the chain rule :
 
       d                     1
       --( arcsin(u) ) = -------------.u'
       dx                   _______
                           |      2
                          \| 1 - u

Analogous

 
       d                   - 1
       --( arccos(u) ) = -------------.u'
       dx                    _______
                            |      2
                           \| 1 - u

       d                     1
       --( arctan(u) ) = -------------.u'
       dx                  1 + u2



       d                    - 1
       --( arccot(u) ) = -------------.u'
       dx                   1 + u2



Examples:

 
       d                   - 1
       --( arccos(x7) ) = ------------ 7 x6
       dx                    ________
                            |      14
                           \| 1 - x

       d                     1
       --( arcsin(1/x)) = ------------- (-1/x2)
       dx                   ___________
                           |          2
                          \| 1 - (1/x)



       d                           1
       --( arctan(sin(2x)) ) = ---------------- (cos(2x)).2
       dx                       1 + sin2(2x)

Rational power of u = f(x)

The n-th root of x

For x > 0 we have
 
        (x(1/n))n=  x
So, these two functions must have the same derivative.
We differentiate both sides.
 
            (1/n) n-1  d    (1/n)
        n.(x     )   . -- (x     ) = 1
                       dx

<=>    ...

        d    (1/n)     1      1/n - 1
<=>     -- (x     ) = ---.  (x        )
        dx             n

The n-th root of u = f(x)

From previous formula we have
 
        d    (1/n)     1      1/n - 1
        -- (u     ) = ---.  (u        ).u'
        dx             n

The square root of u

From previous formula we have
 
        d    (1/2)     1       -1/2
        -- (u     ) = ---.  (u     ).u'
        dx             2
So,
                       1
     D( sqrt(u) ) = --------- u'
                    2.sqrt(u)

Rational power of x

 
        d    (m/n)    d    (1/n) m      (1/n) m-1  1      1/n - 1
        -- (x     ) = -- (x     ) = m.(x     )   . --.  (x        )
        dx            dx                           n

<=>     ...

           m
<=>     = --- . xm/n - 1
           n

A rational power of u = f(x)

From previous formula we have
 
        d    (m/n)      m
        -- (u     ) =  --- . um/n - 1.u'
        dx              n
 
        d    (m/n)      m
        -- (u     ) =  --- . um/n - 1.u'
        dx              n


   d                    1
   -- ( sqrt(u) ) = --------- u'
   dx               2.sqrt(u)

Examples:

 
                          6x
 D( sqrt(3x2+5) ) = --------------
                     2. sqrt(3x2+5)


 D( (sin(3x))4/3 ) = (4/3)(sin(3x))1/3 . cos(3x) . 3

                               cos(x)cos(2x) -2sin(x)sin(2x)
 D( sqrt(sin(x).cos(2x)) ) = --------------------------------
                              2 sqrt(sin(x).cos(2x))



                                2x+1
                      (2x+1). ---------------  - sqrt(x2+x+1). 2
   sqrt(x2+x+1)                2.sqrt(x2+x+1)
 D ------------- = ----------------------------------------------- = ....
    2x + 1                     (2x+1)2



                                     2x                           2x
                    sqrt(x2+4). --------------- - sqrt(x2-2) ---------------
   sqrt(x2-2)                    2.sqrt(x2-2)                 2.sqrt(x2+4)
 D ------------ = ------------------------------------------------------------ = ...
   sqrt(x2+4)                  (x2+4)

Logarithmic functions

See derivative of a logarithmic function

Exponential functions

See derivative of an exponential function

Real power of x

See derivative of a real power of x

uv

See derivative of uv

Partial derivatives

definition of partial derivatives

Suppose we have a function f such that the image depends on several independent variables, say x, y and z.

We write f(x,y,z).

Consider, for a moment, in such function f(x,y,z) y and z as constant, then f(x,y,z) only depends on the variable x. Then, we can calculate the derivative with respect to x.
We note this derivative as

 
        fx'(x,y,z)
Similarly we can calculate
 
        fy'(x,y,z)     and     fz'(x,y,z)

Example of partial derivatives

 
f(x,y,z) = 3 x y - x z + 6 y -4


        fx'(x,y,z) = 3 y - z

        fy'(x,y,z) = 3 x + 6

        fz'(x,y,z) = - x

Extension of the chain rule

Suppose we have a function f(x,y,z) such that x, y and z are not independent but functions of a variable t. Then f is a function of t.

Suppose that x, y and z are differentiable functions of t. We denote the three derivatives for short as x',y' and z'. We can calculate the derivative of f with respect to t with the formula

 
      d
      --- f(x,y,z) = fx'(x,y,z) . x' +  fy'(x,y,z) . y' + fz'(x,y,z) . z'
      dt

The proof of this formula is beyond the scope of this tutorial.

Example:

 
f(x,y,z) = x2 + x.y + z2 + y.z

and  x = 3t ; y = 5 t2 ; z = t3

Then
      d
      --- f(x,y,z) = (2x + y).3 + (x + z). 10t + (2z + y). 3t2
      dt

                   = (6t + 5t2).3 + (3t + t3) .10t + (2t3 + 5t2).3t2

                   = ...

                   = 6 t5 + 25 t4 + 45 t2 + 18 t

Implicit Differentiation

Implicit functions

Example :

We define the function
 
    x + 6
y = -----
    x - 5
If x is not 5, the same function is defined by one the implicit forms
 
y.(x - 5) = x + 6

or

x y = x + 5 y + 6

or

x y - x - 5 y - 6 = 0

Implicit definition of a function

From previous example we see that expressions as
 
        g(x,y) = 0
or      g(x,y) = h(x,y)
can define y implicitly as a function of x.

Implicit differentiation

Concept of implicit differentiation

Suppose that we know that y is implicitly defined as a function of x by the expression
 
        g(x,y) = h(x,y)
If we think y as a function of x, the previous form expresses an identity for all x-values.
Then, the left side is a function of x, and the right side is a function of x and these functions are identical.
Therefore the derivative, with respect to x, of both sides is the same.
 
d g(x, y)   d h(x, y)
--------- = ---------
   dx         dx
This defines implicitly y'.

Example of implicit differentiation

From previous example we know that
 
        x y = x + 5 y + 6 
defines y implicitly as a function of x.
If we think y as a function of x, the previous form expresses an identity for all x-values.
Then, the left side is a function of x, and the right side is a function of x and these functions are identical.
Therefore the derivative, with respect to x, of both sides is the same.
 
        1.y + x.y' = 1 + 5 y' + 0

<=>
        (x-5) y' = 1 - y

<=>
             1 - y
        y' = -----
             x - 5
Previous procedure to calculate y' is called implicit differentiation

Generalization

Previous procedure can be generalized to calculate y' even if the expression
 
        g(x,y) = h(x,y)
defines two or more different y values as a function of x.
Example:
 
 2
y = 4x defines two different y values as a function of x.

Thinking of y as a function of x, we obtain
 
        2 y y' = 4
<=>
        y' = 2/y
This result holds for the two different functions!!

Maximum and minimum values

Relative maximum

We say that a function f(x) has a relative maximum for x = t if and only if there is a strictly positive real number e such that f(x) =< f(t) for all x in ]t-e,t+e[ .
Often the word 'relative' is omitted.

Relative minimum

We say that a function f(x) has a relative minimum for x = t if and only if there is a strictly positive real number e such that f(x) >= f(t) for all x in ]t-e,t+e[ .
Often the word 'relative' is omitted.

Roots of f'(x)

If f'(x) exists in an open interval that contains the value x = t, and suppose that f reaches a maximum for x = t.

Then, f'(t) = 0.


Proof:
 
       f'(x) exists for x = t

             (f(t+h)-f(t))
=>       lim --------------- = a number g
       h->0        h

    We consider the right and left limit separately

            (f(t+h)-f(t))                   (f(t+h)-f(t))
=>      lim --------------- = g  and    lim --------------- = g
        > 0       h                     < 0       h


     Since f(x)  reaches a maximum for x = t , we have

=>           ( g > or =  0 )           and       ( g < or = 0 )

=>           g = 0

=>           f'(t) = 0
In the same way we can prove :

If f'(x) exists in an open interval that contains the value x = t, and suppose that f reaches a minimum for x = t.

Then, f'(t) = 0.


Rolle's theorem

The function f(x) is given If
  • f(a) = f(b)
  • f is continuous in [a,b]
  • f is differentiable in ]a,b[
Then, there is a c-value in ]a,b[ such that f'(c) = 0.

Proof:
If f is constant in [a,b], then the proof is trivial.

Now, suppose f is not constant in [a,b].

Then there is a number d in ]a,b[ such that f(d) is different from f(a) and f(b).

Suppose first that f(d) > f(a). Since f is continuous in [a,b], f attains, according to Weierstrass, a maximum image.

So, there is a c-value in ]a,b[ such that f(c) is maximum and since f is differentiable in ]a,b[, f'(c) = 0.

In the same way, you can prove the theorem for f(d) < f(a).

As soon as at least one of the three conditions is not satisfied, the existence of the c-value is not guaranteed.

Lagrange's theorem

If
  • f is continuous in [a,b]
  • f is differentiable in ]a,b[
Then, there is a c-value in ]a,b[ such that
 
              f(b) - f(a)
     f'(c) = ---------------
                 b - a

C is the graph of y = f(x).
The slope of PQ is ( f(b) - f(a) ) / (b - a).
f'(c) is the slope of the tangent line in point R(c,f(c)).
The theorem says that there is a suitable point R on the curve C such that the tangent line in R is parallel to PQ.

Proof:

The equation of PQ is

 
                 f(b) - f(a)
    y - f(a) =  -------------- (x - a)
                    b - a
<=>
                 f(b) - f(a)
    y = f(a) +  -------------- (x - a)
                    b - a
Now, we consider three functions
 
    f :  x --> y = f(x)

                            f(b) - f(a)
    g :  x --> y = f(a) +  -------------- (x - a)
                               b - a

    h :  x --> y = f(x) - g(x)
It is not difficult to verify that the h(x) satisfies the three conditions of Rolle's theorem.

We apply Rolle's theorem on the function h(x). There is a c-value in ]a,b[ such that h'(c) = 0;

 
    h'(x) = f'(x) - g'(x)

                          f(b) - f(a)
          = f'(x) - 0 -  ------------- .1
                             b - a
There is a c-value in ]a,b[ such that
 
               f(b) - f(a)
     f'(c) -  ------------- = 0
                 b - a

<=>
               f(b) - f(a)
     f'(c) =  ------------
                 b - a

Constant functions

Theorem:
If f'(x) = 0 for all x in [a,b]
Then f(x) is constant in [a,b].

Proof:
Take two values c and d in [a,b]. We'll show that f(c) = f(d).

Since f'(x) exists for each x in [a,b], f'(x) exists in [c,d] and f(x) is continuous in [c,d].
We apply lagrange's-theorem in [c,d].
There is a value e in ]c,d[ such that

 
               f(c) - f(d)
     f'(e) =  -------------
                 c - d

but f'(x) = 0 for all x in [c,d]. So,

       f(c) - f(d)
      ------------- = 0
         c - d

and  f(c) = f(d)

Increasing functions

Theorem:
If f'(x) > 0 for all x in [a,b]
Then f(x) is increasing in [a,b].

Proof:
Take two values c < d in [a,b]. We'll show that f(c) < f(d). Since f'(x) exists for each x in [a,b], f'(x) exists in [c,d] and f(x) is continuous in [c,d].
We apply lagrange's-theorem in [c,d].
There is a value e in ]c,d[ such that

 
               f(c) - f(d)
     f'(e) =  -------------
                 c - d

but f'(x) > 0 for all x in [c,d]. So,

       f(c) - f(d)
      ------------- > 0
         c - d

Since c < d , we have f(c) < f(d).

Decreasing functions

Theorem:
If f'(x) < 0 for all x in [a,b]
Then f(x) is decreasing in [a,b].

(proof similar to previous theorem)

Relative maximum of a differentiable functions.

Let e = a strictly positive real number.
Suppose that a function is differentiable in ]t-e,t+e[ .

If f'(x) > 0 in a suitable interval ]t-e,t[ , the f(x) is increasing at the left side of t.
If f'(x) < 0 in a suitable interval ]t,t+e[ , the f(x) is decreasing at the right side of t.
In that case f(x) has a relative maximum for x = t.

Relative minimum of a differentiable functions.

Let e = a strictly positive real number.
Suppose that a function is differentiable in ]t-e,t+e[ .

If f'(x) < 0 in a suitable interval ]t-e,t[ , the f(x) is decreasing at the left side of t.

If f'(x) > 0 in a suitable interval ]t,t+e[ , the f(x) is increasing at the right side of t.

In that case f(x) has a relative minimum for x = t.

Detection of a relative maximum or relative minimum

In general we detect a relative maximum or relative minimum of f(x) by investigating the sign of f'(x).
f(x) has a maximum = f(t) if f'(x) changes from + to - for x=t.

f(x) has a minimum = f(t) if f'(x) changes from - to + for x=t.


Example

 
f(x) = 3x5- 5x3 then f'(x) = 15x4-15x2= 15 x2.(x-1)(x+1)

Investigation of the sign gives

        x       |       -1      0       1
        ---------------------------------------
        f'(x)   |   +       -       -       +

So there is a maximum for x = -1 and a minimum for x = 1

Applications of differentiation

Rotating light

A rotating light is at a distance of 3m from a long wall. It rotates one revolution per second. At time t = 0 point P is illuminated. What is the speed of light on the wall at the moment that u = pi/3 radians?
 
             
As 1 revolution takes just one second, u = 2.pi.t
The velocity v of the spot on the wall is ds/dt. The relationship between s and the angle u is s = 3 tan(u).
From this we have s = 3 tan (2.pi.t).
 
       ds         2 pi              6pi
  v = ---- = 3 -------------- = ---------------
       dt      cos2(2 pi t)      cos2(2 pi t)

  If  u = 2 pi t =  pi/3   then    v = 24 pi  m/s

Balloon

Gas is escaping from a spherical balloon with a flow rate of 1 cm3/s.
How rapidly reduces the surface of the balloon when the radius is 20 cm.
 
             

  Area of the sphere = A =  4 pi r2
Volume of the sphere = V = (4/3) pi r3
dA/dt = 8 pi r dr/dt (*) DV/dt = 4 pi r2 dr/dt We know that DV/dt = -1 So, 4 pi r2 dr/dt = -1 => dr/dt = -1/(4 pi r2) (**) From (*) and (**) it follows that dA/dt = 8 pi r . ( -1/(4 pi r2) ) = -2/r On the moment that r = 20, dA/dt = -0.1 The area decreases with 0.1 cm2/s when the radius r is 20 cm

Water

A cone-shaped vessel is filled with water. At that time the water surface has a diameter of 4 dm. The height of the cone is 1 dm.
0.1 liter per second flows from the vessel. What is the speed with which the water level falls on the moment that the water level is 0.5 dm.
 
            

  Let h = the water level at a given time t.
  Let r = the radius of water surface at a given time t.

  V = Volume water at a given time t.  V = pi r2 h / 3.

  r and h are functions of t.

  d V/dt = (pi/3) ( 2 r (dr/dt) h + (dh/dt) r2 )    (*)

  But we know that  dV/dt = -0.1

  The figure it follows that: r/h = 2/1 ; so, r = 2h and  dr/dt =2 dh/dt

  (*) becomes:     -0.1 = (pi/3) (12 h2 )(dh/dt)

  At the moment that h = 0.5 we have dh/dt = -0.03

  At the time that the water level is 0.5 dm, the water level falls 3 mm/s.

Snell's law

In an area A is the velocity of a particle d = v1 and in the area B its velocity is v2. The particle d starts at point P1 and should go to point P2 in the shortest possible time. We will find a remarkable feature about the orbit of the particle.
 
            
We choose to axes as shown in the figure.

Let P1(0,a) P2(c,-b) S(x,0)

The total time T to go from A to B is a function of x.

 
  T = |P1 S|/v1 + |S P2|/v2

       sqrt(a2+x2)      sqrt( (c-x)2 + b2 )
    = --------------- + ----------------------
           v1                  v2

  The total time T should be minimum.

  dT         2 x                2 (c-x)(-1)
 --- = ----------------- + --------------------------
  dx    v1 2 sqrt(a2+x2)     v2 2 sqrt((c-x)2 + b2 )

  T is minimum if and only if dT/dx = 0. Then we have


               x                   (c-x)
        ----------------- = --------------------------      (*)
         v1  sqrt(a2+x2)     v2  sqrt((c-x)2 + b2 )

  The angles i and j are marked on the figure.
  In the right triangles on the figure we have

                 x
            -------------- = sin(i)
            sqrt(a2+x2)

and
                (c-x)
           --------------------- = sin(j)
           sqrt((c-x)2 + b2 )

So, the condition (*) becomes

          sin(i)       sin(j)
        --------- =  ---------
            v1          v2

          sin(i)         v1
        --------- =  ---------
          sin(j)         v2

And we find Snell's law !

Concavity

Concave upward

If, in an interval, f"(x) > 0 , then f'(x) is increasing.
Then the slope of the tangent line is increasing with x.
We say that f(x) is concave upward in that interval.

If, in an interval, f"(x) > 0 then f(x) is concave upward in that interval.

Concave downward

If, in an interval, f"(x) < 0 , then f'(x) is decreasing. Then the slope of the tangent line is decreasing with x. We say that f(x) is concave downward in that interval.

If, in an interval, f"(x) < 0 then f(x) is concave downward in that interval.

Points of inflection

Example:
 
f(x) = 3x5- 5x3 then  f"(x) = 15(x4 - x2) = 30x(2x2- 1)

Investigation of the sign gives

        x       |       -sqrt(1/2)      0       sqrt(1/2)
        ----------------------------------------------------
        f"(x)   |   -              +        -              +

So,
for x < -sqrt(1/2) the graph is concave downward.
for -sqrt(1/2) < x < 0  the graph is concave upward.
for 0 < x < sqrt(1/2) the graph is concave downward.
for x > sqrt(1/2) the graph is concave upward.
The points where the concavity changes sign are the points with
x = -sqrt(1/2) ; x = 0 ; x = sqrt(1/2).
These points are called points of inflection.
The points where the concavity changes sign are the points of inflection.

Theorem of Cauchy

If f(x) and g(x) are continuous in [a,b] and if f'(x) and g'(x) exist in ]a,b[ and have no common roots in ]a,b[ , then there is a value c in ]a,b[ such that
 
         f(b) - f(a)     f'(c)
        ------------- = -------
         g(b) - g(a)     g'(c)

Proof:
 
Denote
         f(b) - f(a)
        ------------- = K              (1)
         g(b) - g(a)

Then f(b) - f(a) - K (g(b) - g(a)) = 0
We create the function f(x) - K g(x).
This function is continuous in [a,b] and the derivative exists in ]a,b[. On this function, we can use Lagrange's theorem .

There is a value c in ]a,b[ such that

 
    f(b) - K g(b) - ( f(a) - K g(a) ) = (b-a).(f'(c) - K g'(c))

=>                      0  = (f'(c) - K g'(c))
If g'(c) = 0 then f'(c) = 0 and this is impossible because f'(x) and g'(x) have no common roots in ]a,b[. So, g'(c) is not 0 and then
 
     f'(c)
    ------ = K                      (2)
     g'(c)
From (1) and (2), we have that there is a value c in ]a,b[ such that
 
         f(b) - f(a)     f'(c)
        ------------- = -------
         g(b) - g(a)     g'(c)

L'Hospitals rule and special limits

Theorem 1

 
If ( lim f(x) = lim g(x) = 0 )
      a          a

          f'(x)
and  lim ------- = can be found
      a   g'(x)


Then
             f(x)         f'(x)
        lim ------ = lim -------
         a   g(x)     a   g'(x)


Proof :
 
                          f'(x)
First, suppose that  lim ------- = finite = A
                     a    g'(x)
Since f'(x) exists in the environment of a, f(x) is continuous in the environment of a and
 
    lim f(x) = f(a)
     a
But we have also that

    lim f(x) = 0
     a
Therefore f(a) = 0.
Similarly g(a) = 0.
With Cauchy's theorem, we can write in the same environment of a
 
    f(x)     f(x) - f(a)     f'(c)
    ----- = ------------- = -------  with c between x and a.
    g(x)     g(x) - g(a)     g'(c)
If x --> a , c --> a.
Now we take the limit of both sides for x --> a
 
        f(x)         f'(c)
   lim ------ = lim ------- = A
    a   g(x)     a   g'(c)


And, in this first case, the theorem is proved.

                        f'(x)
Now, suppose that  lim ------- = + infinity
                     a  g'(x)

Then
           g'(x)
      lim ------- = +0
        a  f'(x)


and appealing on the first case

         g(x)         g'(c)
    lim ------ = lim ------- = +0
     a   f(x)     a   f'(c)
So,
        f(x)         f'(c)
   lim ------ = lim ------- = + infinity
    a   g(x)     a   g'(c)

(similar for - infinity)

Theorem 2

 
If ( lim f(x) = lim g(x) = infinite )
      a          a

          f'(x)
and  lim ------- = can be found
      a   g'(x)


Then
             f(x)         f'(x)
        lim ------ = lim -------
         a   g(x)     a   g'(x)


Proof:

 
                          f'(x)
First, suppose that  lim ------- = finite = A
                     a    g'(x)

Consider the identity for all x

    f(x)     f(x) - f(b)    1 - g(b)/g(x)
    ----- = -------------. ---------------
    g(x)     g(x) - g(b)    1 - f(b)/f(x)
Since f'(x) exists in the environment of a, f(x) is continuous in the environment of a and similarly for g(x).

With Cauchy's theorem, previous identity becomes, with b and x in the same environment of a :

 
    f(x)      f'(c)    1 - g(b)/g(x)
    ----- = ------- . ---------------  with c between x and a.
    g(x)      g'(c)    1 - f(b)/f(x)
If x --> a , c --> a.
Now we take the limit of both sides for x --> a
 
        f(x)         f'(c)    1 - 0
   lim ------ = lim -------. ------ = A
    a   g(x)     a   g'(c)    1 - 0

And, in this first case, the theorem is proved.
                        f'(x)
Now, suppose that  lim ------- = + infinity
                     a  g'(x)

Then
           g'(x)
      lim ------- = +0
        a  f'(x)


and appealing on the first case

         g(x)         g'(c)
    lim ------ = lim ------- = +0
     a   f(x)     a   f'(c)
So,
        f(x)         f'(c)
   lim ------ = lim ------- = + infinity
    a   g(x)     a   g'(c)

(similar for - infinity)

Limits of irrational functions and L'Hospitals rule

L'Hospitals rule may be useful when calculating limits of irrational functions. However, there are cases where this rule is not useful at all. We give some examples.

Special limits - examples

L'Hospitals rule is frequently used in the following limits.
In these examples the properties and the derivatives of exponential and logarithmic functions are used.

Solved problems

 

You can find solved problems about differentiation and limits using this link

You can find solved problems about maximum, minimum and inflection points using this link

You can find solved problems about irrational functions using this link

 




Topics and Problems

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