Lesson 17 (3.9)

Lesson 17 (3.9)

Today:

• Derivatives of logarithmic functions
• Logarithmic Differentiation

Office Hours: MWF 1:30 – 2:30pm

Announcements:

• Exam 2: Monday, 10/19, 8pm – 9pm
• Final Exam: Tuesday, 12/15, 10:30am – 12:30pm

Review: Properties of Logarithms

• \( x = e^y \quad \text{then} \quad y = \ln x \)
• \( \log_b mn = \log_b m + \log_b n \)
• \( \log_b m/n = \log_b m - \log_b n \)
• \( \log_b m^r = r \log_b m \)
• \( b^{\log_b x} = x \)
• \( \log_b b^x = x \)

Derivative of \( y = \ln x \), \( x > 0 \)

\[ y' = \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h} \]

Not easy to compute

instead: use \( y = \ln x \Rightarrow x = e^y \)
and implicit diff.

\[ x = e^y \] \[ \frac{d}{dx} x = \frac{d}{dx} e^y \] \[ 1 = e^y \cdot \frac{dy}{dx} \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x} \]

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

For \(x < 0\), \(\ln x\) is not defined but \(\ln(-x)\) is defined.

Find \(\frac{d}{dx}(\ln(-x))\), for \(x < 0\)

\(u = -x > 0\)

\[ \frac{d}{dx}(\ln(-x)) = \frac{d}{dx}\ln u = \frac{d\ln u}{du} \cdot \frac{du}{dx}, \quad u > 0 \] \[ = \frac{1}{u} \cdot \frac{du}{dx}, \quad u = -x > 0 \] \[ = \frac{1}{-x} \cdot \frac{d}{dx}(-x) = \frac{1}{-x} \cdot -1 \] \[ = \frac{1}{x} \] \[ \frac{d}{dx}\ln|x| = \frac{1}{x}, \quad x \neq 0 \]

eg: \( \frac{d}{dx} \left[ \ln(e^x) \right] \)

CAN Apply Chain Rule, but its much easier if we recognize that \( \ln e^x = x \)

\[ \frac{d}{dx} \left[ \ln e^x \right] = \frac{d}{dx} [x] = 1 \]

Example: \(\frac{d}{dx}[\ln(\sin^4 x)]\)

\[\ln b^r = r \ln b\]

\[\frac{d}{dx}[\ln(\sin^4 x)] = \frac{d}{dx}(4 \ln(\sin x)) = 4 \cdot \frac{d}{dx}[\ln(\sin x)]\]

Apply Chain Rule   \(u = \sin x\)

\[= 4 \cdot \frac{d}{dx} \ln u\]

\[= 4 \cdot \frac{d}{du} \ln u \cdot \frac{du}{dx}\]

\[= 4 \cdot \frac{1}{u} \cdot \frac{du}{dx}\]

\[= 4 \cdot \frac{1}{\sin x} \frac{d}{dx} \sin x = \frac{4 \cos x}{\sin x}\]

Logarithmic Properties:

\[ \ln\left(\frac{m}{n}\right) = \ln m - \ln n \] \[ \ln(mn) = \ln m + \ln n \]

Example:

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

Example

\[ y = x^x \]

\[ y' = x \cdot x^{x-1} \quad \text{\bfseries\Large ✗} \]

Cannot apply power rule when you have a function in power.

\[ \ln y = \ln x^x = x \ln x \]

\[ \frac{d}{dx} [\ln y] = \frac{d}{dx} [x \ln x] \]

\[ \frac{1}{y} \cdot y' = 1 \cdot \ln x + x \cdot \frac{d}{dx} \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]

\[ y' = y [\ln x + 1] = x^x [\ln x + 1] \]

Logarithmic Differentiation

\[ y = [f(x)]^{g(x)} \]

1. take \(\ln\) on both sides

   \[ \ln y = \ln [f(x)^{g(x)}] = g(x) \ln(f(x)) \]

2. take \(\frac{d}{dx}\) on both sides

   \[ \frac{d}{dx} \ln y = \frac{d}{dx} [g(x) \ln(f(x))] \]

3. Apply chain Rule on LHS, product rule on RHS.

   \[ \frac{1}{y} y' = \frac{d}{dx} [g(x) \ln(f(x))] \]

4. solve for \(y'\)

   \[ y' = y \left[ \frac{d}{dx} (g(x) \ln(f(x))) \right] \]

eg: \( y = x^{\sin x} \)

have \( \sin x \) in the power \( \leadsto \) take \( \ln \) on both sides

\[ \ln y = \ln(x^{\sin x}) = \sin x \ln x \]

\( \frac{d}{dx} \) on both sides

\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\sin x \ln x) \] \[ \frac{1}{y} \cdot y' = \cos x \ln x + \sin x \cdot \frac{1}{x} \] \[ y' = y \left[ \cos x \ln x + \sin x \cdot \frac{1}{x} \right] = x^{\sin x} \left[ \cos x \ln x + \frac{\sin x}{x} \right] \]

Example: \( y = \frac{(x^5+3)^7}{(x^3+9)(4x+9)^{1/2}} \), find \( y' \).

Can use Quotient Rule to solve, but Logarithmic differentiation might make it simpler

\[ \ln y = \ln \left[ \frac{(x^5+3)^7}{(x^3+9)(4x+9)^{1/2}} \right] \] \[ = \ln \left( [x^5+3]^7 \right) - \ln \left[ (x^3+9)(4x+9)^{1/2} \right] \] \[ = 7\ln(x^5+3) - \ln(x^3+9) - \frac{1}{2}\ln(4x+9) \] \[ \frac{d}{dx} \ln y = 7 \cdot \frac{d}{dx} \ln(x^5+3) - \frac{d}{dx} \ln(x^3+9) - \frac{1}{2}\ln(4x+9) \] \[ \frac{1}{y} y' = 7 \cdot \frac{1}{x^5+3} \cdot 5x^4 - \frac{1}{x^3+9} \cdot 3x^2 - \frac{1}{2} \cdot \frac{1}{4x+9} \cdot 4 \] \[ y' = y \left[ \frac{35x^4}{x^5+3} - \frac{3x^2}{x^3+9} - \frac{2}{4x+9} \right] = \frac{(x^5+3)^7}{(x^3+9)(4x+9)^{1/2}} \left[ \frac{35x^4}{x^5+3} - \frac{3x^2}{x^3+9} - \frac{2}{4x+9} \right] \]

eg: \( y = b^x \), \( b > 0 \)

\[ \ln y = \ln b^x = x \ln b \] \[ \frac{d}{dx} \ln y = \frac{d}{dx} (x \ln b) \] \[ \frac{1}{y} \cdot y' = \ln b \leadsto y' = y \ln b = b^x \ln b \] \[ \frac{d}{dx} b^x = b^x \ln b, \quad b > 0 \]

Verify:

\[ \frac{d}{dx} e^x = e^x \cdot \ln e = \underline{\underline{e^x}} \]

eg: \( y = \log_b x \), \( b > 0 \)

using properties of log & exponential

\[ x = b^y \] \[ \frac{d}{dx} x = \frac{d}{dx} b^y \] \[ 1 = \frac{d}{dy} b^y \cdot \frac{dy}{dx} = b^y \ln b \cdot \frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{1}{b^y \ln b} = \frac{1}{x \ln b} \] \[ \frac{d}{dx} \log_b x = \frac{1}{x \ln b}, \quad b > 0 \]

eg:

\[ y = \log_{\sqrt{7}}(\tan^3 x) = 3 \log_{\sqrt{7}}(\tan x) \] \[ \begin{aligned} y' &= 3 \cdot \frac{d}{dx}\left[\log_{\sqrt{7}}(\tan x)\right] = 3 \frac{d}{dx} \log_{\sqrt{7}} u \qquad (u = \tan x) \\[10pt] &= 3\left(\frac{d}{du} \log_{\sqrt{7}} u\right) \cdot \left(\frac{du}{dx}\right) \\[10pt] &= 3 \cdot \frac{1}{u \ln 7} \cdot \frac{du}{dx} \\[10pt] &= 3 \cdot \frac{1}{(\tan x)\ln 7} \cdot \frac{d\tan x}{dx} = \frac{3\sec^2 x}{(\tan x)(\ln 7)} \end{aligned} \]