..Exponential
functions
..Logarithmic
functions
..The number e
..Differentiation of
logarithmic functions
..Derivative
of a real power of x..Derivative
of uv Exponential functions
Definition
Take a > 0 and not equal to 1 . Then, the function defined byf : R -> R : x -> axis called an exponential function with base a.
Graph and properties
Let f(x) = an exponential function with a > 1.Let g(x) = an exponential function with 0 < a < 1.
From the graphs we see that
- The domain is R
- The range is the set of strictly positive real numbers
- The function is continuous in its domain
- The function is increasing if a > 1 and decreasing if 0 < a < 1
- The x-axis is a horizontal asymptote
Logarithmic functions
Definition and basic properties
Take a > 0 and not equal to 1 . Since the exponential functionf : R -> R : x -> axare either increasing or decreasing, the inverse function is defined. This inverse function is called the logarithmic function with base a. We write
loga (x)
So,
|
loga(x) = y <=> ay = x |
From this we see that the domain of the logarithmic function is the set of strictly positive real numbers, and the range is R.
Example:
log2(8) = 3 ; log3(sqrt(3)) = 0.5 ;
From the definition it follows immediately that
for x > 0 we have aloga(x) = x
and
for all x we have loga(ax) = x
|
Graph
Let f(x) = a logarithmic function with a > 1.Let g(x) = a logarithmic function with 0 < a < 1.
From the graphs we see that
- The range is R
- The domain is the set of strictly positive real numbers
- The function is continuous in its domain
- The function is increasing if a > 1 and decreasing if 0 < a < 1
- The y-axis is a vertical asymptote
Properties
In the next 3 properties, all logarithmic functions have base a > 0. For convenience, I don't write this base a.- log(x.y) = log(x) + log(y)
Proof :Let log(x.y) = u then au = x.y (1) Let log(x) = v then av = x (2) Let log(y) = w then aw = y (3) From (1) , (2) and (3) au = av . aw => au = av + w => u = v + wSo,log(x.y) = log(x) + log(y)
- In the same way you can prove that
log(x/y) = log(x) - log(y)
- For each real number r we have :
log(xr ) = r.log(x)
Change the base of a logarithmic function
Sometimes it is very useful to change the base of a logarithmic function.Theorem: for each strictly positive real number a and b, different from 1, we have
1
loga(x) =( -------) . logb(x)
logb(a)
|
Proof:
We'll prove that
logb(a) . loga(x) = logb(x)
Let logb(a) = u then bu = a (1)
Let loga(x) = v then av = x (2)
Let logb(x) = w then bw = x (3)
From (2) and (3) we have
av = bw
Using (1)
bu.v = bw
So,
u.v = w
=> logb(a) . loga(x) = logb(x)
The number e
A special limit concerning the derivative of an exponential function
We try to calculate the derivative of the exponential function
f(x) = ax
Appealing on the definition of the derivative, we can write
(f(x+h)-f(x))
f'(x) = lim ---------------
h->0 h
ax+h - ax
= lim ------------
h->0 h
ax (ah - 1)
= lim -----------
h->0 h
(since ax is constant with respect to h )
(ah - 1)
= ax . lim -----------
h->0 h
Now,
(ah - 1)
lim ----------- is a constant depending on the value of the base a.
h->0 h
It can be proved that there is a unique value of a, such that this limit
is 1. This very special value of a is called e.So,
(eh - 1)
lim ----------- = 1
h->0 h
The number e as a limit
The expression
(eh - 1)
lim ----------- = 1
0 h
means that for very very small values of h
eh - 1 is approximately h
<=> eh is approximately h +1
<=> e is approximately (1 + h)1/h
So,
e = lim (1 + h)1/h = 2.718 28...
0
|
Or, if we say that t = 1/h
e = lim (1 + 1/t)t = 2.718 28...
infty
|
Definition of ln(x)
The logarithmic function with base number e is noted ln(x). So,
|
loge(x) = ln(x) |
Differentiation of logarithmic functions
Derivative of a logarithmic function
In this section, all logarithmic functions have base a. For convenience, I don't write this base number.
Let f(x) = log(x) , then
(f(x+h)-f(x))
f'(x) = lim ---------------
h->0 h
(log(x+h)-log(x))
<=> f'(x) = lim -------------------
h->0 h
log( (x+h)/x )
<=> f'(x) = lim -------------------
h->0 h
1
<=> f'(x) = lim --- . log( (x+h)/x )
h->0 h
<=> f'(x) = lim log( (x+h)/x )1/h
h->0
<=> f'(x) = lim log( (x+h)/x )1/h
h->0
<=> f'(x) = lim log(1 + h/x)1/h
h->0
<=> f'(x) = lim log((1 + h/x)x/h )1/x
h->0
<=> f'(x) = lim (1/x).log(1 + h/x)x/h
h->0
<=> f'(x) =(1/x). lim log(1 + h/x)x/h
h->0
<=> f'(x) =(1/x). lim log(1 + h/x)x/h
x/h->0
<=> f'(x) =(1/x).log lim (1 + h/x)x/h
x/h->0
<=> f'(x) =(1/x).log(e)
<=> f'(x) =(1/x).ln(e)/ln(a)
<=> f'(x) =(1/x)/ln(a)
1
<=> f'(x) = ----------
x. ln(a)
Important cases
Let u be a differentiable function of x.
d 1
-- loga(x) = ----------
dx x. ln(a)
d 1
-- loga(u) = ---------- . u'
dx u. ln(a)
d 1
-- ln(x) = ---
dx x
d 1
-- ln(u) = ---.u'
dx u
|
Derivative of an exponential function
Let f(x) = ax, then loga(ax ) and x are identical functions.
Hence, the derivative of both functions is the same.
So,
1
---------- .(ax )' = 1
ax .ln(a)
d
<=> ---(ax ) = ax .ln(a)
dx
Important corollaries
Let u be a differentiable function of x.
d
---(ax ) = ax .ln(a)
dx
d
--(ex ) = ex
dx
d
--(au ) = au .ln(a).u'
dx
d
--(eu ) = eu .u'
dx
|
Derivative of a real power of x
Let f(x) = xr with r any real number.
xr = er.ln(x)
=>
d
--(xr) = er.ln(x).(r.ln(x))'
dx
= xr.r.(1/x)
= r.xr-1
Thus,
For any real number r, we have d --(ur) = r.ur-1.u' dx |
Derivative of uv
Let u = f(x) and v = g(x), then
uv = ev.ln(u)
d
--(uv) = ev.ln(u).(v.ln(u))'
dx
= uv . (v' ln(u) + v.(1/u).u'
= v uv-1 u' + uv.ln(u).v'
d --(uv) = v uv-1 u' + uv.ln(u).v' dx |
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