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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

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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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