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									Chemistry 1S                             Calculus I                           Dr Paul May




4.12 Types of Stationary Point

If xsp is the stationary point, then if we consider points either side of xsp, there are 4
types of behaviour of the gradient.

                    x < xsp    x = xsp      x > xsp           t.p.type

              (i)       +ve      zero        -ve             maximum
             (ii)       -ve      zero        +ve             minimum
            (iii)       +ve      zero        +ve             inflection
            (iv)        -ve      zero        -ve             inflection


Stationary points like (iii) and (iv), where the gradient doesn't change sign produce S-
shaped curves, and the stationary points are called points of inflection.




           Minimum             Maximum                 +ve p.o.i      -ve p.o.i



4.13 How to determine if a stationary point is a max, min or point
of inflection

The rate of change of the slope either side of a turning point reveals its type. But a
                                                                                   dy
rate of change is a differential. So all we need to do is differentiate the slope,
                                                                                   dx
with respect to x. In other words we need the 2nd differential, or

        d  dy                      d2 y
             , more usually called                   (“dee-2-y by dee-x-squared”)
        dx  d x                    dx 2

Examples

1.     y(x) = 9x2 - 2

                dy                               d2 y
                   = 18x                              = 18
                dx                               dx 2



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Chemistry 1S                              Calculus I                           Dr Paul May


                      1
2.     y(x) = 4x5 -
                      x

        dy          1                             d2 y          2
           = 20x4 + 2                                2
                                                       = 80x3 - 3
        dx         x                              dx           x

3.     p = 3q3 - 4q2 + 6

        dp                                        d2 p
           = 9q2 - 8q                                  = 18q - 8
        dq                                        dq 2



4.14 Rules for stationary points

                                                  d2 y
       i)      At a local maximum,                     = -ve
                                                  dx 2

                                                  d2 y
       ii)     At a local minimum,                     = +ve
                                                  dx 2

                                                d2 y
       iii)    At a point of inflection,             = 0,
                                                dx 2
               and we must examine the gradient either side of the turning point to
               find out if the curve is a +ve or -ve p.o.i.

Examples

1.     Taking the same example as we used before:

       y(x) = x3 - 3x + 1        (a cubic function – you should be expecting 2 t.p.s)

        dy
           = 3x2 - 3       stationary points at ( -1, 3) and (1, -1)
        dx

        d2 y
             = 6x
        dx 2
                                         d2 y
At stationary point (-1,3), x = -1, so        = -6,    so this point is a maximum.
                                         dx 2

                                          d2 y
At stationary point (1, -1), x = +1, so        = +6,    so this point is a minimum
                                          dx 2

So we can finally sketch the curve:




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Chemistry 1S                                            Calculus I                                   Dr Paul May



               max at (-1,3)
                                                        y
                  3

                  2

                  1

                  0                                                                           x
                 -1

                 -2                                                                  min at (1,-1)
                 -3
                      -3          -2          -1            0            1       2        3




2.     y = x3 + 8                           (a cubic function – you should be expecting 2 t.p.s)

        dy
           = 3x2,              which is equal to zero at the turning point(s).
        dx

       If 3x2 = 0, x = 0,               and so y = +8, so the t.p. is at (0,8).

       [The fact that we’ve found only one turning point for a cubic function (for
       which you would normally expect two) should tell you either you’ve done it
       wrong, or that the t.p. might be ‘unusual’].

               d2 y
       Now          = 6x
               dx 2

                                       d2 y                                            dy
So, at the t.p. (0,8), i.e when x = 0,    2
                                            = 0, so we have a point of inflection. But
                                       dx                                              dx
                                                      dy
is +ve either side of this point (e.g. at x = +1,         = +3 i.e. increasing, at x = -1,
                                                      dx
 dy
    = +3, also increasing), so the curve has a positive point of inflection.
 dx
                       40
                                                                point of inflection at (0,8)

                       30                           y

                       20


                       10
                                                                                          x
                           0


                       -10
                          -3           -2      -1           0        1       2        3       4




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Chemistry 1S                                 Calculus I                               Dr Paul May


3.      Where are the turning point(s), and does it (or they) indicate a max or min
        in the function p(q) = 4 - 2q - 3q2 ? [How many t.p.s do we expect?]

                 dp
                    = -2 - 6q, which at the turning point = 0
                 dq
                so -2 - 6q = 0,    6q = -2, so q = - 1 , and p = 4 1
                                                      3             3


        We have one t.p. at ( - 1 , 4 1 ).
                                3     3



         d2 p
              = -6, which is –ve, so the t.p. is a maximum.
         dq 2
(This is consistent with what we said earlier, that for quadratics if the x2 term is -ve,
we have a -shaped parabola).

                       turning point at ( -1/3, 13/3 )
                          6
                                                       p
                          4


                          2
                                                                        q
                          0


                         -2


                         -4


                         -6
                              -3   -2        -1            0       1    2




4.15 More Complicated Differentiation Rules

So far, we’ve only dealt with simple polynomial functions. But in chemistry we’re
likely to meet a whole bunch of other, more complicated functions, which we also
need to differentiate. For many of these, the ‘magic formula’ is insufficient, so we
need to learn 3 more rules, and recognise when to apply them. These rules are: the
Product Rule, the Quotient Rule, and the Function of a Function rule.


4.15A Product Rule

Say we have functions like:

     y(x) = (3x2 – 2x +3)(4x4 +6x - 9);      or        x2 ln x ;       or   3x sinx

where the function y(x) can be written as two other functions multiplied together, e.g.
in the second example we have x2×lnx. We name the first function u(x) and the
second one v(x), so that




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Chemistry 1S                              Calculus I                         Dr Paul May


       y(x) = u(x)  v(x)

then the differential of y(x) is given by the Product Rule:

                   dy      dv      du
                      = u.    + v.
                   dx      dx      dx
In words: ‘The first function multiplied by the differential of the second, plus the
second function multiplied by the differential of the first’.

Examples

1.     y(x) = (x2 + 2)(x + 1)

                                        du
       let u = x2 + 2,          so that    = 2x
                                        dx
                                        dv
       let v = x + 1,           so that    =1
                                        dx

            dy      dv      du
       so      = u.    + v.          = (x2 + 2).1 + (x + 1).2x
            dx      dx      dx

               = 3x2 + 2x + 2

(Note: in this case we can double check this is correct by expanding out the brackets)

                                     dy
       y(x) = x3 + x2 + 2x + 1,         = 3x2 + 2x + 2
                                     dx

                         1
2.     y(x) = x3 (3 -      + 3x2)
                         x
                        
               u         v

                dy        1              1
                   = x3 ( 2 + 6x) + (3 -   + 3x2).3x2
                dx       x               x

                            1   1
3.     () = (2 + 2)(      + 2)
                             
                             
                     u        v

        dφ                1   2      1  1
           = (2 + 2)( - 2 - 3 ) + ( + 2 )(2+2)
        dλ                          




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Chemistry 1S                                    Calculus I                                        Dr Paul May




4.15B Quotient Rule

Sometimes we come across rational functions, where one function is divided by
another. Examples include:

                    x 1                                   x2                           log10 x
        y ( x)                        or      y ( x)            or         y ( x) 
                   3x  9                                 sin x                          x 1

Again, we call them u and v, making sure that u is on the top of the fraction, so that:

                   u( x )
       y(x) =             ,
                   v( x)

To differentiate this, we use the Quotient Rule:

                                 du   dv 
                               v   u  
                             =   2  
                          dy      dx     dx
                          dx          v

Example
                              x2  1                                       du
                   y(x) =                     so that u = x2 + 1       and    = 2x
                              2x  3                                       dx
                                                                           dv
                                              and     v = 2x + 3       and    =2
                                                                           dx

                              dy   (2 x  3)(2 x )  ( x 2  1)(2)
                                =
                              dx            (2 x  3) 2

You may need to simplify/tidy up the answer:

                              dy   (4 x 2  6 x)  (2 x 2  2)          2x2  6x  2
                                 =                                 
                              dx           (2 x  3) 2                   (2 x  3) 2

There will be lots more examples of how to use both the Product Rule and the
Quotient Rule in the tutorial/workshops.




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