Chemistry 1S
Calculus I
Dr Paul May
5.3 Variants of the Exponential Function
In most applications the exponent is not simply x, but some function of x, or the exponent
is multiplied by some other function of x.
Notation:
We normally write the exponential function as y = ex, but if the exponent
is a function (e.g. x2 + 1), it's often easier to write it as y = exp(x2 + 1), which is
2
equivalent to e x 1 . (Note - This ‘exp’ is not the same as ‘EXP on your calculator, which
refers to 10x)
Examples
y = e-bx,
y = ebx,
y = e 2x
2
1
or
 ax
b
,
y = exp(2x2 - 1)
-bx
y = 5ex,
y= e
y = 5x.e
y = (x2 - 1)e  x ,
2
,
etc.
5.4 Pre-exponential factors and Time Constants
A generic exponential function is: y(x) = Ae-bx,
where A and b are constants.
The number or expression in front of the exponential (in the above example this is A) is
called the ‘pre-exponential factor’. You can see that when x=0, y=A, so this is actually
the intercept on the y-axis. You’ll come across this in kinetics, where the Arrhenius
equation shows that the rate constant k of a chemical reaction depends on temperature, T,
via:
k  Ae

Ea
RT
In this case, when the activation barrier Ea for the reaction is zero, so that all collisions
result in a reaction, then k = A. So the pre-exponential factor A has a real meaning here, it
is the rate at which molecules collide.
38
Chemistry 1S
Calculus I
Dr Paul May
Any constant, b, that multiplies to x inside the exponential has the effect of changing the
rate of decay (or increase) of y (see plot, above). For this reason it is often called a ‘rate
constant’, a ‘decay constant’, or a ‘time constant’. For example, in a first-order reaction,
the concentration of a reactant A depends upon time via:
[A]t(t) = [A]0 e-kt
where [A]t means concentration of A at time, t, [A]0 is concentration at time t = 0 (i.e. the
starting concentration), and k is the first-order rate constant.
5.5 Differentiating ex
The reason why ex is so important is that it is the only function which doesn't change
when you differentiate it.
i.e.
if
dy
= ex
dx
y = ex,
Examples
1.
y = 5ex,
dy
= 5ex
dx
2.
y = 3ex + 2,
dy
= 3ex
dx
3.
y = ebx  chain rule with u = bx,
and
So
dy
dy
du
=
×
=
du
dx
dx
du
=b
dx
dy
y = eu,
= eu
du
bebx
39
Chemistry 1S
Calculus I
Dr Paul May
Alternatively, using the sequential step rule:
dy
= ebx × b
dx
 
diff. of diff of
(...)
e
bx
Or as a general rule:
d bx
(e ) = bebx
dx

dy
= - e-x
dx
4.
y = e-x,
5.
y = exp(x2 + 1),
6.
y = x.e -bx. This requires the product rule
 
u v
dy
= exp(x2 + 1) × 2x
dx


diff of
diff of
exp(...)
x2+1
dy
dv
du
=u
+v
= x(be-bx) + e-bx.(1)
dx
dx
dx
7.
= (1 - bx)e-bx
What is the gradient of the curve y = 5e-2x + 1 at the point x = 2?
dy
=
dx
10
y
8
and at x = 2, slope = .
6
4
2
0
-1
40
0
1
2
3
4
5
x
Chemistry 1S
Calculus I
Dr Paul May
6. Logarithmic Functions
If we have the relationship
y = ax
then there must be the inverse relationship such that
x = f (y).
We call the function, f (y), the ‘logarithm to base a’.
x = loga(y)
valid only for y>0
There are 2 types of logarithm in common use:
a) Common logs have base 10 and are written log10x
b) Natural logs have base e and are written
logex or ln x
So, if
y = ex,
then ln y = x
y = 10x,
then log10 y = x
6.1 Laws of Logarithms
1.
ln A + ln B = ln (AB)
2.
ln A - ln B = ln (A/B)
3.
ln Ax = x.ln A
Examples
1.
ln 2 + ln 3 = 1.792 = ln 6
2.
ln x + ln (x2+1) = ln {x.(x2+1)} = ln (x3+x)
3.
ln 6 - ln 3 = 0.693 = ln 2
41
Chemistry 1S
Calculus I
4.
 x 1 
ln (x+1) - ln (3 – x2) = ln 

 3  x2 
5.
 x ( x  3) 
ln x + ln (x+3) - ln (x2+4) = ln  2

 x 4 
6.
ln (x2 + 1)3 = 3ln (x2 + 1)
7.
ln (x2+3)x+1 = (x+1) ln (x2+3)
8.
 1
ln   =
 x
ln(x -1)
Dr Paul May
 important
= - ln x
Numerical Example: when x =2,
9.
ln
( x  1) 2
4x
or
=
ln
32
8
=
ln
9
8
= 2ln (x+1) - ln (4x) =
=
0.118
2ln 3 - ln 8
=
0.118
6.2 The Differential of ln x
It can be shown that, if
y = ln x,
1
dy
=
x
dx
We can now use this, together with the Product, Chain and Sequential Rules to find the
differentials of log functions.
Examples
1.
y = ln (ax + b)
dy
1
=
a =
dx
(ax  b)


Differential of Diff. of
ln (1/(...) )
(...)
a
(ax  b)
42
Chemistry 1S
2.
Calculus I
y = ln (2x) + 3x2
=
dy  1  2
=
 6x
.
d x  2 x  2 x 1/ 2
3.
ln (2x½) + 3x2
=
1
 6x
2x
Where are the turning points in y = ln(x) – x ?
1
dy
= 1
x
dx
1
 1 = 0,
x
i.e., when x =1, and y = -1.
So the slope = 0 when
1
d2 y
= - 2 , which at (1,-1) is -ve,
2
x
dx
so the t.p. is a maximum.
43
Dr Paul May