Calculus I: exponential function
logarithmic function and derivative
1. Derivative: exponential functions
1.1. Definitions
1.1.1. Derivative of ex
d/dx(ex) = ex
By definition,
d/dx(ex) = | lim( ex + Δx - ex)/Δx = 0/0 = Undetermined |
| Δx → 0 |
= ex | lim (eΔx - 1)/Δx |
| Δx → 0 |
We know :
lim (ex - 1)/x = 1
x → 0
Therefore:
dex/dx = ex
d/dx(ex) = ex
1.1.2. Derivative of eg(x)
let:
f(x) = y , and g(x) = u, so y = eu.
According to the chain rule formula:
dy/dx = (dy/du).(du/dx)
= eu . d/dx(g(x)) = eu . g'(x) =
f(x) . g'(x) = eg(x) . g'(x)
d/dx(eg(x)) = g'(x) . eg(x)
1.1.3. Derivative of bg(x)
We make use of the change of base rule for
the exponential function bg(x):
(bg(x) = eln b g(x)
d/dx (bg(x)) = d/dx ( eln bg(x)) =
d/dx (eln b . g(x)) = d/dx (ln b . g(x)) . (eln b . g(x)) =
ln b . d/dx (g(x)) . (eln b . g(x)) = ln b . d/dx (g(x)) . (bg(x)) =
Therefore:
d/dx (bg(x)) = ln b . d/dx (g(x)) . (bg(x))
d(bg(x))/dx = ln b . g'(x). bg(x)
1.2. Examples
a) f(x) = e4x
f'(x) = (4x)'. e4x = 4 e4 x
b) f(x) = 3x ex2
f'(x) = (3x)' ex2 + 3x (ex2)'
= 3 ex2 + 3x ((x2)' ex2) =
= 3 ex2 + 3x (2 x ex2) =
= 3 ex2 + 6 x2 ex2 =
= ex2 (3 + 6 x2)
c)
f(x) = 23x
d(23x)/dx = ln 2 . 3. 23x = 3 (ln 2) 23x
d)
f(x) = (1/2)x + 4 = - (ln 2) (1/2)x + 4
2. Derivative: logarithmic functions
2.1. Definitions
2.1.1. Derivative of ln(x)
Let y = ln(x) then ey = x . So
d/dx(ey) = d/dx(x) = 1
dy/dx . (ey) = d/dx(x) = 1
Therefore:
dy/dx = 1/ey = 1/x
Hence:
d ln(x) /dx = 1/x
d ln(x) /dx = 1/x
2.1.2. Derivative of ln g(x)
Let y = ln u with g(x) = u
Using the chain rule for derivative, we have:
dy/dx = (dy/du).(du/dx) = (1/u).(du/dx) = (1/g(x)).(dg(x)/dx)
Therefore:
d ln(g(x))/dx = (1/g(x)).(dg(x)/dx)
d ln(g(x))/dx = (1/g(x)).(dg(x)/dx) = g'(x)/g(x)
2.1.3. Derivative of logbg(x)
We make use of the change of base for
the logarithmic function logbg(x):
logbg(x) = logeg(x) / logeb = ln g(x)/ ln b
d/dx[logbg(x)] =
d/dx[ln g(x)/ ln b] =
(1/ln b) . d/dx[ln g(x)] = (1/ln b) . d/dx[g(x)] /g(x)
Therefore:
d/dx[logbg(x)] = (1/ln b) . d/dx(g(x)) /g(x)
d/dx[logbg(x)] = (1/ln b) . d (g(x))/dx /g(x) =
g'(x)/ln b. g(x)
2.1.4. Logarithmic derivatives
Logarithmic derivative is used when the derivative
of some functions is so long or so complicated to perform. These
functions generally include products, ratios, or exponents.
This technique do the job very quickly.
Let y = f(x). f(x) is the "complicated" function.
The method is then:
-
Take the ln of both sides:
-
ln y = ln f(x)
-
Set the derivative of both sides:
-
(1/y).dy/dx = f'(x)/f(x)
-
Once f'(x)/f(x) is solved, we write the expression of
dy/dx.
-
dy/dx = y [f'(x)/f(x)]
The logarithmic derivative is used also to differentiate
expressions of the form:
y = f(x)g(x)
That is when the base of the exponantial f(x) and the
exponent g(x) depend on the independent variable x.
In the general case where
y = uv, the logarithmic derivative gives:
ln y = v ln u , so
(1/y)dy/dx = (dv/dx) ln u + v (1/u)(du/dx)
Therefore:
dy/dx = y[(dv/dx) ln u + v (1/u)(du/dx)] = uv[(dv/dx) ln u + v (1/u)(du/dx)] =
uv (dv/dx) ln u + v uv - 1(du/dx)
Hence:
The derivative of y = uv is:
dy/dx = uv (dv/dx) ln u + v uv - 1(du/dx)
y = uv
dy/dx = uv (dv/dx) ln u + v uv - 1(du/dx)
2.1.5. Absolute value: logarithmic derivatives
Using the relationship:
d ln(f(x))/dx = (1/f(x)).(df(x)/dx) = f'(x)/f(x)
We write:
d ln |f(x)|/dx =
d ln f(x)/dx = (1/f(x)).(df(x)/dx) =
f'(x)/f(x) if f(x) > 0
= d ln (- f(x))/dx = (1/(- f(x))).(df(x)/dx) =
f'(x)/f(x) if f(x) < 0
In both case, the result is f'(x)/f(x).
Therefore:
d ln |f(x)|/dx = (1/f(x)).(d f(x)/dx) = f'(x)/f(x)
d ln |f(x)|/dx = (1/f(x)).(dg(x)/dx) = f'(x)/f(x)
Now, we want to derive an absolute value:
y = |f(x)|
We have:
dy/dx =
d f(x)/dx if f(x) > 0
- d f(x)/dx if f(x) < 0
Using the logarithmic derivative, we have:
ln y = ln |f(x)|. So
(1/y)(dy/dx) = f'(x)/f(x)
dy/dx = y . f'(x)/f(x)
That is:
d|f(x)|/dx = |f(x)|. f'(x)/f(x)
d|f(x)|/dx = |f(x)|. f'(x)/f(x)
2.2. Examples
a)
f(x) = ln(x) /x
f'(x) = (x ln'(x) - (x)' ln (x))/x2 = (x (1/x) - 1. ln (x))/x2 =
(1 - ln (x))/x2
b)
f(x) = ln (2 x + 3)
f'(x) = (2 x + 3)'/(2 x + 3) = 2/(2 x + 3)
c)
f(x) = log3 (2 x + 1)
f'(x) = (1/ln 3) . (2 x + 3)'/(2 x + 3) = (2/ln 3)/(2 x + 3)
d)
f(x) = log (x)
f'(x) = (1/ln 10) . (x)'/x = (1/ln 10)/x
e)
f(x) = x2 (x + 1)2/(x3 + 1)
y = f(x)
ln y = ln f(x) =
ln [x2 (x + 1)2/(x3 + 1)] =
ln x2 + ln (x + 1)2 - ln (x3 + 1) =
2 ln x + 2 ln (x + 1) - ln (x3 + 1)
Taking the derivatives yields:
(1/y)dy/dx = 2 (1/x) + 2. 1/(x + 1) - (3x2)/(x3 + 1) =
2/x + 2/(x + 1) - 3x2/(x3 + 1)
Then:
dy/dx = y [2/x + 2/(x + 1) - 3x2/(x3 + 1)] =
x2 (x + 1)2/(x3 + 1) .
[2/x + 2/(x + 1) - 3x2/(x3 + 1)]
Therefore:
dy/dx =
x2 (x + 1)2/(x3 + 1) .
[2/x + 2/(x + 1) - 3x2/(x3 + 1)]
Expression that can be reduced.
f)
f(x) = |2 x - 1|
ln f(x) = ln |2 x - 1|
d ln f(x)/dx = d ln |2 x - 1|/dx = d ln |2 x - 1|/dx = d ln (2 x - 1)/dx = 2/(2 x - 1)
g)
y = |4x - 5|
ln y = ln |4x - 5|. So
(1/y)(dy/dx) = 4/(4 x - 5)
d|4x - 5|/dx = 4 |4x - 5|/(4x - 5)
3. Exercises
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