Download original notes
PAGE 1

3.9 Derivatives of Logarithmic Functions

NOT on exam 2

We know

\[ \frac{d}{dx} e^x = e^x \quad \text{and} \quad \frac{d}{dx} e^u = e^u \frac{du}{dx} \]
Together with implicit differentiation, we can now find derivatives of logarithmic functions.

\[ y = \ln x \]

is equivalent to \[ x = e^y \]

Now differentiate \[ x = e^y \] implicitly:

\[ \frac{d}{dx}(x) = \frac{d}{dx}(e^y) \]

Note: \( y \) is an implicit function of \( x \), so chain rule applies.

\[ 1 = e^y \left( \frac{dy}{dx} \right) \]

We want this: \( \frac{dy}{dx} \)

\[ \frac{dy}{dx} = \frac{1}{e^y} \quad \text{but } y = \ln x \] \[ = \frac{1}{e^{\ln x}} = \frac{1}{x} \]

Summary of Rules

So, \[ \frac{d}{dx} \ln x = \frac{1}{x} \quad x \neq 0, x > 0 \]

And \[ \frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx} \]

PAGE 2

Example 1

\[ y = \ln(\underbrace{\tan x}_{u}) \quad \text{using } \frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx} \] \[ \frac{dy}{dx} = \frac{1}{\tan x} \frac{d}{dx} \tan x = \frac{1}{\tan x} \sec^2 x \]
\[ \frac{\sec^2 x}{\tan x} \]

Example 2

\[ y = \ln(\underbrace{\ln(12x)}_{u}) \] \[ \frac{dy}{dx} = \frac{1}{\ln(12x)} \underbrace{\frac{d}{dx} \ln(12x)}_{\text{chain rule again}} \] \[ = \frac{1}{\ln(12x)} \cdot \frac{1}{12x} \cdot \frac{d}{dx}(12x) \] \[ = \frac{1}{\ln(12x)} \cdot \frac{1}{12x} \cdot 12 \]
\[ = \frac{1}{x \ln(12x)} \]
PAGE 3

Example: Logarithmic Differentiation

\[ y = \ln \left[ (x^3 + 7)^{4\pi} \right] \]

Rule:

\[ \frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx} \]

Applying the chain rule where \( u = (x^3 + 7)^{4\pi} \):

\[ \frac{dy}{dx} = \frac{1}{(x^3 + 7)^{4\pi}} \cdot \frac{d}{dx} (x^3 + 7)^{4\pi} \]

Chain rule again:

\[ = \frac{1}{(x^3 + 7)^{4\pi}} \cdot 4\pi (x^3 + 7)^{4\pi - 1} \cdot \frac{d}{dx} (x^3 + 7) \]
\[ = \frac{1}{(x^3 + 7)^{4\pi}} \cdot 4\pi (x^3 + 7)^{4\pi - 1} \cdot 3x^2 \]
\[ = \frac{12x^2\pi (x^3 + 7)^{4\pi - 1}}{(x^3 + 7)^{4\pi}} = 12x^2\pi (x^3 + 7)^{-1} \]
\[ = \frac{12x^2\pi}{x^3 + 7} \]
PAGE 4

Alternative Way: Use Property of Log First

Property: \( \ln a^b = b \ln a \)

\[ y = \ln (x^3 + 7)^{4\pi} \quad \text{use } \ln a^b = b \ln a \]

Rewrite as:

\[ y = 4\pi \ln (x^3 + 7) \]

Now differentiate:

\[ \frac{dy}{dx} = 4\pi \cdot \frac{d}{dx} \ln (x^3 + 7) \]

Differentiating \( \ln(u) \):

\[ = 4\pi \cdot \frac{1}{x^3 + 7} \cdot \frac{d}{dx} (x^3 + 7) = 4\pi \cdot \frac{1}{x^3 + 7} \cdot 3x^2 \]
\[ = \frac{12x^2\pi}{x^3 + 7} \]
PAGE 5

Example: Differentiation of Logarithmic Functions

\[ y = \ln \left( \frac{x+1}{x-1} \right) \]

Option 1: Direct Differentiation

Using the chain rule with the formula:

\[ \frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx} \]

Let \( u = \frac{x+1}{x-1} \). Then:

\[ \frac{dy}{dx} = \frac{1}{\frac{x+1}{x-1}} \cdot \frac{d}{dx} \left( \frac{x+1}{x-1} \right) \]

Applying the quotient rule to the second term:

\[ \frac{dy}{dx} = \frac{x-1}{x+1} \cdot \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} \]
\[ = \frac{1}{x+1} \cdot \frac{-2}{x-1} = \frac{-2}{(x+1)(x-1)} \]

Option 2: Using Logarithm Properties

Log property: \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \)

Rewrite the original function:

\[ y = \ln \left( \frac{x+1}{x-1} \right) \]
\[ y = \ln(x+1) - \ln(x-1) \]

Now differentiate term by term:

\[ \frac{dy}{dx} = \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) - \frac{1}{x-1} \cdot \frac{d}{dx}(x-1) \]
PAGE 6
\[ = \frac{1}{x+1} - \frac{1}{x-1} \]

Verification

Is it equal to \( \frac{-2}{(x+1)(x-1)} \)?

\[ \frac{1}{x+1} \cdot \frac{x-1}{x-1} - \frac{1}{x-1} \cdot \frac{x+1}{x+1} = \frac{(x-1) - (x+1)}{(x+1)(x-1)} = \frac{-2}{(x+1)(x-1)} \]

General Rule

In general, it is always easier to handle more simpler log functions than to deal with fewer but more complicated log functions.

PAGE 7

We can take advantage of log properties even if the logs don't appear initially

Logarithmic Differentiation

Example

\[ y = (x^4 + 2)^3 (x^5 + 1)^6 \]

Option 1: Leave as is, use product + chain rules

\[ \frac{dy}{dx} = (x^4 + 2)^3 \cdot \frac{d}{dx}(x^5 + 1)^6 + (x^5 + 1)^6 \cdot \frac{d}{dx}(x^4 + 2)^3 \]

Note: The terms \( (x^5 + 1)^6 \) and \( (x^4 + 2)^3 \) are of the form \( u^n \), requiring the chain rule.

\[ = (x^4 + 2)^3 \cdot 6(x^5 + 1)^5 \cdot 5x^4 + (x^5 + 1)^6 \cdot 3(x^4 + 2)^2 \cdot 4x^3 \]\[ = \dots \text{ (simplify)} \]
PAGE 8

Option 2: Logarithmic Differentiation

\[ y = (x^4 + 2)^3 (x^5 + 1)^6 \]

Take \( \ln \) on both sides:

\[ \ln y = \ln \left[ (x^4 + 2)^3 (x^5 + 1)^6 \right] \]\[ = \ln (x^4 + 2)^3 + \ln (x^5 + 1)^6 \]

Log Properties Used:

  • \( \ln(ab) = \ln(a) + \ln(b) \)
  • \( \ln(a^b) = b \ln a \)
\[ \ln y = 3 \ln (x^4 + 2) + 6 \ln (x^5 + 1) \]

Goal: Make whatever following \( \ln \) as simple as you can.

Differentiate implicitly, solve for \( \frac{dy}{dx} \):

\[ \frac{d}{dx} \ln y = \frac{d}{dx} [3 \ln (x^4 + 2)] + \frac{d}{dx} [6 \ln (x^5 + 1)] \]\[ \frac{1}{y} \frac{dy}{dx} = 3 \cdot \frac{1}{x^4 + 2} \cdot 4x^3 + 6 \cdot \frac{1}{x^5 + 1} \cdot 5x^4 \]

Want this!

\[ \frac{1}{y} \left( \frac{dy}{dx} \right) = \frac{12x^3}{x^4 + 2} + \frac{30x^4}{x^5 + 1} \]\[ \frac{dy}{dx} = y \left( \frac{12x^3}{x^4 + 2} + \frac{30x^4}{x^5 + 1} \right) \]
\[ = (x^4 + 2)^3 (x^5 + 1)^6 \left( \frac{12x^3}{x^4 + 2} + \frac{30x^4}{x^5 + 1} \right) \]
PAGE 9

Logarithmic Differentiation Example

Example: \( y = x^x \)

\( \frac{dy}{dx} = ? \)

Note: \( \frac{d}{dx} x^n = nx^{n-1} \) only if exponent is a constant.

Take \( \ln \) on both sides:

\( \ln y = \ln x^x \) Using property: \( \ln(a^b) = b \ln a \)

\( \ln y = (x)(\ln x) \)

Differentiate implicitly:

\( \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x \cdot (1) \) Product rule

\( \frac{1}{y} \frac{dy}{dx} = 1 + \ln x \)

\( \frac{dy}{dx} = y(1 + \ln x) = \)\( x^x(1 + \ln x) \)

PAGE 10

General Exponential Differentiation

We know \( \frac{d}{dx} e^x = e^x \), but \( \frac{d}{dx} a^x \neq a^x \) if \( a \neq e \).

e.g., \( \frac{d}{dx} 2^x \neq 2^x \). But what is it?

\( y = a^x \)

\( \frac{dy}{dx} = ? \)

Use logarithmic differentiation:

\( \ln y = \ln a^x \)

\( \ln y = x \cdot \ln a = (\ln a)x \) (\(\ln a\) is a constant)

Differentiate implicitly:

\( \frac{1}{y} \frac{dy}{dx} = \ln a \)

\( \frac{dy}{dx} = y \ln a = a^x \ln a \)

General Rule

\( \frac{d}{dx} a^x = a^x \ln a \)

Check: if \( a = e \)

Does it match what we know?

\( \frac{d}{dx} a^x = a^x \ln a \)

If \( a = e \), then \( \frac{d}{dx} e^x = e^x \ln e \)

\( \frac{d}{dx} e^x = e^x \)

Similarly,

\( \frac{d}{dx} \log_a x = \frac{1}{x \ln a} \)