Exponential and Logarithmic Functions
Differential and Integral Calculus
Prof. Lahcen Laayouni
School of science and engineering
El Akhawayn University
Exponential functions: Differentiation and integration
Derivatives of logarithmic functions
Wednesday, November 21st , 2007
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Outline
1
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarithmic functions
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Definition 1: (Derivative of exponential function)
The derivative of ex is D[ex ] = ex . We say that ex is invariant with
respect to the derivative.
Definition 2: (Derivative of ax )
Since ax = ex ln a , using the chain rule, we obtain
D(ax ) = D(ex ln a ) = (ln a)ex(ln a) = (ln a)ax .
Definition 3: (Derivative of ag(x) and eg(x) )
Using the chain rule, we find that
D[ag(x) ] = (ln a)ag(x) g ′ (x) ,
and
D[eg(x) ] = eg(x) g ′ (x) .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Example 1:
Find the derivative of f (x) = 3e2x
2 −5
.
Solution:
Let u(x) = 2x 2 − 5 then f (x) = 3eu(x) . Using the chain rule, we
obtain
D(f (x)) = 3u ′ (x)eu(x) = 3(4x)e2x
2 −5
= 12xe2x
2 −5
.
Example 2:
Find the derivative of g(x) = 3
√
x 3 +2x
.
Solution:
Let u(x) =
√
x3
+ 2x , so
u ′ (x)
=
Prof. Lahcen Laayouni
1√
3x 2 +2
2 x 3 +2x
=
√
2
1 (3x +2) x 3 +2x
2
x 3 +2x
Differential and Integral Calculus
.
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Solution: (Continue)
Thus D[3u(x) ] = (ln 3)3u(x) u ′ (x) , so.
D[3
√
x 3 +2x
√
√
ln 3 (3x 2 + 2) x 3 + 2x
x 3 +2x
]=
.
3
2
x 3 + 2x
Example 3:
√
Find the derivative of h(x) = ex x .
Solution:
Using the product rule, we get
√
1
h′ (x) = ex . √ + xex
2 x
√
√ 2 x
ex
ex
x
√ + e x. √ = √ (1 + 2x) .
=
2 x
2 x
2 x
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Theorem 1: (Integration rules for exponential functions)
If u is a differentiable function of x , then
Z
ex dx = ex + C,
and
Z
eu du = eu + C .
Example 4:
Evaluate
Z
xe−x
2 +1
dx .
Solution:
Let u = −x 2 + 1 , so that du = −2xdx , thus xdx = −du/2 , then
Z
Z
1
1
1
2
−x 2 +1
xe
dx = −
eu du = − eu + C = − e−x +1 + C .
2
2
2
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Theorem 2: (Derivative of ln(x) )
To find the derivative of ln(x) , we use the definition of the derivative,
we obtain
x +h
ln
d (ln(x))
ln(x + h) − ln(x)
x
= lim
= lim
dx
h→0
h
h→0
h
1
h
h 1/h
= lim ln 1 +
= lim ln 1 +
.
h→0 h
x
h→0
x
Now, let
h
x
=
1
m
, then
1
h
=
m
x
. As h → 0 , m → ∞ , so that
1/x
1 m/x
1 m
d (ln(x))
= lim ln 1 +
= lim ln 1 +
.
m→∞
m→∞
dx
m
m
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Derivative of ln(x) (continue)
Since ln(x) is a continuous function, then we can bring the limit inside
of the naturel logarithm to get
1/x
d (ln(x))
1 m
= ln lim 1 +
m→∞
dx
m
Since
lim
m→∞
1
1+
m
m
=e .
therefore, the derivative of ln(x) , becomes
d (ln(x))
= ln
dx
lim
m→∞
1
1+
m
Prof. Lahcen Laayouni
m 1/x
= ln(e1/x ) =
Differential and Integral Calculus
1
.
x
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Derivative of loga (x)
Consider the general logarithmic function, f (x) = loga (x) , we solve for
x
f (x) = loga (x) ⇒ af (x) = x .
Differentiating both sides of the last equation with respect to x , we
obtain
(ln(a))af (x) f ′ (x) = 1 ⇒ (ln(a))x f ′ (x) = 1 ,
Finally, dividing both sides of this equation by (ln(a))x to get
f ′ (x) =
1
.
(ln(a))x
Derivative of logarithmic functions
a) D [loga (x)] =
1
.
(ln(a))x
Prof. Lahcen Laayouni
b) D [ln(x)] =
1
.
x
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Example 5:
Find the derivative of the following functions
1− f (x) = ln(2x) . 2− g(x) = log(x) . 3− log5 (3x 2 ) for x > 0 .
Solution:
1- Using the properties of logarithms and rule of derivatives, we obtain
f ′ (x)
=
=
d
d
(ln(2x)) =
(ln(2) + ln(x ))
dx
dx
d
d
1
1
(ln(2)) +
(ln(x)) = 0 + = .
dx
dx
x
x
2- Recall that log(x) = log10 (x) , so that
d
d
1
(log(x)) =
(log10 (x)) =
dx
dx
(ln(10))x
3- We have log5 (3x 2 ) = log5 (3) + log5 (x 2 ) = log5 (3) + 2 log5 (x) .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Solution:(continue)
3- Thus
d
d
d
2
(log5 (3x 2 )) =
(log5 (3)) +
(2 log5 (x)) =
.
dx
dx
dx
ln(5)(x)
Derivative of the form loga g(x)
Using the chain rule, with g(x) = g(x) and f (x) = loga (x) , we obtain
d
d
1 g ′ (x)
loga g(x) =
[f (g(x))] = f ′ (g(x)).g ′ (x) =
.
.
dx
dx
ln(a) g(x)
If a = e , then
d
g ′ (x)
ln(g(x)) =
.
dx
g(x)
Example 6:
Find 1− D[ln(2x 2 + 1)] .
Prof. Lahcen Laayouni
2− D[log5 (4x 2 − 3x + 1)] .
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Solution:
1- Let g(x) = 2x 2 + 1 , so that g ′ (x) = 4x , then
D[ln(2x 2 + 1)] =
g ′ (x)
4x
.
=
g(x)
2x 2 + 1
2- Let g(x) = 4x 2 − 3x + 1 , so that g ′ (x) = 8x − 3 , then
D[log5 (4x 2 − 3x + 1)] =
1 g ′ (x)
1
8x − 3
=
.
ln(5) g(x)
ln(5) 4x 2 − 3x + 1
Remark 1:
Consider f (x) = ln(−x) , where x < 0 . Applying the chain rule with
g(x) = −x , g ′ (x) = −1 , gives f ′ (x) =
g ′ (x)
−1
1
=
= .
g(x)
−x
x
So the derivative of f (x) = ln(−x) is the same as that of y = ln(x) .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Derivative of loga |x| , loga |g(x)| , ln |x| , and ln |g(x)|
1- D [loga |x|] =
2- D [ln |x|] =
1
,
ln(a)x
1
,
x
D [loga |g(x)|] =
D [ln |g(x)|] =
1 g ′ (x)
.
.
ln(a) g(x)
g ′ (x)
.
g(x)
Example 7:
Find the derivatives of each function.
1- f (x) = ln |ax| , where a is a non-zero constant.
2- g(x) = 4x 3 ln(| − 2x 2 + 5|) .
Solution
1− Let g(x) = ax , so g ′ (x) = a , and
Prof. Lahcen Laayouni
d
g ′ (x)
a
1
(ln |ax|) =
=
= .
dx
g(x)
ax
x
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Solution: (continue)
2− Using the product rule, we obtain
d
′
3
2
ln(| − 2x + 5|) + 12x 2 ln(| − 2x 2 + 5|)
g (x) = (4x )
dx
−4x
2 ln(| − 2x 2 + 5|)
+
12x
−2x 2 + 5
=
(4x 3 )
=
16x 4
+ 6x 2 ln(2x 2 − 5)2 .
2x 2 − 5
Example 8:
Find the derivative of f (x) = x
√
x
, for x > 0 .
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Solution:
Using the fact that x = eln(x) , then we can rewrite the function f (x) as
√
√x
f (x) = eln(x)
= e x ln(x) .
Applying the chain rule and the derivative of logarithmic functions gives
√
d √
d h √x ln(x) i
e
= e x ln(x)
f ′ (x) =
x ln(x) .
dx
dx
Next we use the product rule, to obtain
√
√ 1
1
′
x
ln(x)
√ ln(x) + x
f (x) = e
,
x
2 x
√
√
ln(x)
1
x x
x
√ +√
= x
= √ (ln(x) + 2)
2 x
x
2 x
Prof. Lahcen Laayouni
Differential and Integral Calculus
Exponential and Logarithmic Functions
Diff. and Int. of exponential functions
Derivatives of logarith
Exponential and Logarithmic Functions
Theorem 3: (Log Rule For Integration)
Let u be a differentiable function of x
Z
1
dx = ln |x| + C,
and
x
Prof. Lahcen Laayouni
Z
1
du = ln |u| + C .
u
Differential and Integral Calculus