The exponential function (Sect. 7.3) The inverse of the logarithm

The exponential function (Sect. 7.3)

The inverse of the logarithm. Derivatives and integrals. Algebraic properties.

The inverse of the logarithm

Remark: The natural logarithm ln : (0, ) R is a one-to-one

function, hence invertible.

y

y = exp (x)

Definition

The exponential function, exp : R (0, ), is the inverse of the natural logarithm, that is,

e

y = ln (x)

1

1

e

x

exp(x) = y x = ln(y ).

y = x

Remark: Since ln(1) = 0, then exp(0) = 1.

Since ln(e) = 1, then exp(1) = e.

The inverse of the logarithm

Remark: Since ln em/n

m

m

= ln(e) = , then holds

n

n

m exp

= em/n

n

The exponentiation of a rational number is the power function.

The exponentiation is a way to extend the power function from rational numbers to irrational numbers.

Definition

For every x R we denote ex = ln-1(x) = exp(x).

Example

Find x solution of e3x+1 = 2.

Solution: Compute ln on both sides,

ln e3x+1

1 = ln(2) 3x + 1 = ln(2) x = ln(2) - 1 .

3

The exponential function (Sect. 7.3)

The inverse of the logarithm. Derivatives and integrals. Algebraic properties.

Derivatives and integrals

Remark: The derivative of the exponential is the same exponential. Theorem (Derivative of the exponential)

(a) For every x R holds ex = ex . (b) For every differentiable function u holds eu = eu u .

Proof:

(a)

ln ex = x

d ln ex = 1 dx

1 ex

ex

=1

We conclude that ex = ex .

(b) Chain rule implies

eu

de u =u

eu = eu u .

du

Remark: In particular: eax = a eax , for a R.

Derivatives and integrals

Remark:

Part (a) of the Theorem can be proven with the formula

f -1 (x) =

1

, for f -1(x) = ex . Indeed,

f f -1(x)

ex =

1

d ln

1

,

(y ) =

ex = ex.

ln ex

dy

y

Example

Find y for y (x ) = e(3x2+5).

Solution: We use all the well-known derivation rules, y = e(3x2+5) = e(3x2+5) (3x 2 + 5) y = 6x e(3x2+5).

Derivatives and integrals

Example

Find y for y (x ) = esin(3x2) ln(x 2 + 1).

Solution: We start with the product rule,

y = esin(3x2) ln(x 2 + 1) + esin(3x2) ln(x 2 + 1)

y = esin(3x2) sin(3x 2)

ln(x 2

+

1)

+

e sin(3x 2 )

(x 2

1 +

1)

(x 2

+

1)

y

=

e sin(3x 2 )

cos(3x2) (6x) ln(x2

+

1)

+

e sin(3x 2 )

(2x ) (x2 + 1) .

y

= 2x esin(3x2)

3

cos(3x 2)

ln(x 2

+

1)

+

(x 2

1 +

1)

.

Derivatives and integrals

Remark: The derivation rule for the exponential implies that its

antiderivative is

eax dx = eax + c. a

Example

/4

Find I =

e3 sin(2x) cos(2x ) dx .

0

Solution: Use the substitution u = 3 sin(2x), du = 6 cos(2x) dx.

I=

/4

e3 sin(2x) cos(2x ) dx =

1 eu du

1 =

eu

1

.

0

0

66

0

Since,

I = 1 (e1 - e0),

we obtain

1 I = (e - 1).

6

6

Derivatives and integrals

Example

Find I = 3x e2x2 sin e2x2 dx .

Solution: Recall that e2x2 = (2x2) e2x2 = 4x e2x2.

Therefore, use the substitution u = e2x2, since du = x e2x2 dx, 4

du

3

3

I = 3 sin(u) = sin(u) du = - cos(u)

4

4

4

Substitute back the original unknown,

I

3 =-

cos

e 2x 2

.

4

Derivatives and integrals

Example

Find the solution to the initial value problem y (x) = 18 e3x , y (0) = 1, y (0) = 2

Solution: We first find y , integrating the equation above, y (x) dx = 18 e3x dx + c y = 18 e3x + c = 6 e3x + c. 3

The initial condition fixes c, 2 = y (0) = 6 e0 + c c = -4 y (x) = 6 e3x - 4.

We now need to integrate one more time.

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