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Calculating derivatives of natural logarithms

Calculate the following derivatives:

  1. d d x ln ( 5 x 3 2 )
  2. d d x ( ln ( 3 x ) ) 2

We need to apply the chain rule in both cases.

  1. d d x ln ( 5 x 3 2 ) = 15 x 2 5 x 3 2
  2. d d x ( ln ( 3 x ) ) 2 = 2 ( ln ( 3 x ) ) · 3 3 x = 2 ( ln ( 3 x ) ) x
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Calculate the following derivatives:

  1. d d x ln ( 2 x 2 + x )
  2. d d x ( ln ( x 3 ) ) 2
  1. d d x ln ( 2 x 2 + x ) = 4 x + 1 2 x 2 + x
  2. d d x ( ln ( x 3 ) ) 2 = 6 ln ( x 3 ) x
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Note that if we use the absolute value function and create a new function ln | x | , we can extend the domain of the natural logarithm to include x < 0 . Then ( d / ( d x ) ) ln | x | = 1 / x . This gives rise to the familiar integration formula.

Integral of (1/ u ) du

The natural logarithm is the antiderivative of the function f ( u ) = 1 / u :

1 u d u = ln | u | + C .

Calculating integrals involving natural logarithms

Calculate the integral x x 2 + 4 d x .

Using u -substitution, let u = x 2 + 4 . Then d u = 2 x d x and we have

x x 2 + 4 d x = 1 2 1 u d u 1 2 ln | u | + C = 1 2 ln | x 2 + 4 | + C = 1 2 ln ( x 2 + 4 ) + C .
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Calculate the integral x 2 x 3 + 6 d x .

x 2 x 3 + 6 d x = 1 3 ln | x 3 + 6 | + C

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Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

Properties of the natural logarithm

If a , b > 0 and r is a rational number, then

  1. ln 1 = 0
  2. ln ( a b ) = ln a + ln b
  3. ln ( a b ) = ln a ln b
  4. ln ( a r ) = r ln a

Proof

  1. By definition, ln 1 = 1 1 1 t d t = 0 .
  2. We have
    ln ( a b ) = 1 a b 1 t d t = 1 a 1 t d t + a a b 1 t d t .

    Use u -substitution on the last integral in this expression. Let u = t / a . Then d u = ( 1 / a ) d t . Furthermore, when t = a , u = 1 , and when t = a b , u = b . So we get
    ln ( a b ) = 1 a 1 t d t + a a b 1 t d t = 1 a 1 t d t + 1 a b a t · 1 a d t = 1 a 1 t d t + 1 b 1 u d u = ln a + ln b .
  3. Note that
    d d x ln ( x r ) = r x r 1 x r = r x .

    Furthermore,
    d d x ( r ln x ) = r x .

    Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have
    ln ( x r ) = r ln x + C

    for some constant C . Taking x = 1 , we get
    ln ( 1 r ) = r ln ( 1 ) + C 0 = r ( 0 ) + C C = 0 .

    Thus ln ( x r ) = r ln x and the proof is complete. Note that we can extend this property to irrational values of r later in this section.
    Part iii. follows from parts ii. and iv. and the proof is left to you.

Using properties of logarithms

Use properties of logarithms to simplify the following expression into a single logarithm:

ln 9 2 ln 3 + ln ( 1 3 ) .

We have

ln 9 2 ln 3 + ln ( 1 3 ) = ln ( 3 2 ) 2 ln 3 + ln ( 3 −1 ) = 2 ln 3 2 ln 3 ln 3 = ln 3 .
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Use properties of logarithms to simplify the following expression into a single logarithm:

ln 8 ln 2 ln ( 1 4 ) .

4 ln 2

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Defining the number e

Now that we have the natural logarithm defined, we can use that function to define the number e .

Definition

The number e is defined to be the real number such that

ln e = 1 .

To put it another way, the area under the curve y = 1 / t between t = 1 and t = e is 1 ( [link] ). The proof that such a number exists and is unique is left to you. ( Hint : Use the Intermediate Value Theorem to prove existence and the fact that ln x is increasing to prove uniqueness.)

This figure is a graph. It is the curve y=1/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1 and to the right at x=e. The area is labeled “area=1”.
The area under the curve from 1 to e is equal to one.

The number e can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series ). Its approximate value is given by

e 2.71828182846 .

The exponential function

We now turn our attention to the function e x . Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by exp x . Then,

Questions & Answers

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The nanotechnology is as new science, to scale nanometric
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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