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Section 6.2 Derivatives and Integrals Involving Log Functions

An introduction to derivatives and integrals involving logrithmic functions.

Subsection 6.2.1 Derivatives of Logrithmic Functions

If \(f(x) = \ln(x)\text{,}\) then what is \(f'(x)\text{?}\)

\begin{align*} f'(x) \amp = \lim\limits_{h\to 0} \dfrac{f(x+h) - f(x)}{h} \\ \amp = \lim\limits_{h\to 0} \dfrac{\ln(x+h) - \ln(x)}{h} \\ \amp = \lim\limits_{h\to 0} \dfrac{\ln(\dfrac{x+h}{x})}{h} \\ \amp = \lim\limits_{h\to 0} \dfrac{\ln(\dfrac{x}{x}+\frac{h}{x})}{h} \\ \amp = \lim\limits_{h\to 0} \dfrac{\ln\left(1+\dfrac{h}{x}\right)}{h} \\ \amp = \lim\limits_{h\to 0} \dfrac{1}{h} \ln\left(1 + \dfrac{h}{x}\right) \\ \amp = \lim\limits_{v\to 0} \dfrac{1}{vx} \ln(1+v) \\ \amp = \lim\limits_{v\to 0} \dfrac{1}{v} \cdot \dfrac{1}{x} \ln(1+v) \\ \amp = \dfrac{1}{x} \lim\limits_{v\to 0} \dfrac{1}{v} \ln(1+v) \\ \amp = \dfrac{1}{x} \lim\limits_{v\to 0} \ln\left((1+v)^{\dfrac{1}{v}}\right)\\ \amp = \dfrac{1}{x} \ln\left(\lim\limits_{v\to 0} (1+v)^{\dfrac{1}{v}}\right) \\ \amp = \dfrac{1}{x} \ln(e) \\ \amp = \dfrac{1}{x} (1) \\ f'(x)\amp = \dfrac{1}{x} \end{align*}

Definition 6.2.1. General Definition of the Derivative of ln(x).

A more general form of \(\dfrac{d}{dx}(\ln(x)) = \dfrac{1}{x}\) is \(\dfrac{d}{dx}(\ln(u)) = \dfrac{1}{u} \cdot \dfrac{du}{dx}\text{.}\)

Since \(\dfrac{d}{dx}(\ln(x)) = \dfrac{1}{x}\text{,}\) \(\displaystyle \int \dfrac{1}{x} dx = \ln(x) + c\text{.}\)

Power Rule for Integrals.

Recall, the power rule for integrals:

\begin{equation*} \int x^n dx = \dfrac{x^{n+1}}{n+1} + c \end{equation*}

* The only exception to the power rule is when \(n = -1\text{.}\)

\begin{equation*} \int \dfrac{1}{x} dx = \int x^{-1} dx \neq \dfrac{x^0}{0} + c \end{equation*}

Subsection 6.2.2 Logrithmic Differentiation

How would we find \(\dfrac{dy}{dx}\) if \(y = \dfrac{(x^4 + 7)\sqrt{x}}{\sin(x)}\text{?}\)

This can be done in three steps:

  1. take the natural log of both sides,

  2. take the derivative of both sides with respect to x,

  3. multiply both sides times y.

Note 6.2.2.
Make sure to simplify your equation between each step.
Example 6.2.1.

Using the equation above:

\begin{align*} y \amp = \dfrac{(x^4 + 7) \sqrt{x}}{\sin(x)}\\ \ln(y) \amp = \ln \left( \dfrac{(x^4 + 7) \sqrt{x}}{ \sin(x)} \right) \\ \amp = \ln((x^4 + 7)\sqrt{x})-\ln(\sin(x)) \\ \amp = \ln(x^4 + 7) + \ln(\sqrt{x})-\ln(\sin(x)) \\ \amp = \ln(x^4 + 7) + \frac{1}{2}\ln(x)-\ln(\sin(x))\\ \frac{1}{y} \cdot \dfrac{dy}{dx} \amp = \dfrac{1}{x^4+7} \cdot 4x^3 + \frac{1}{2} \cdot \frac{1}{x} - \dfrac{1}{\sin(x)} \cdot \cos(x) \\ \dfrac{dy}{dx} \amp = \left( \dfrac{1}{x^4+7} \cdot 4x^3 + \frac{1}{2} \cdot \frac{1}{x} - \dfrac{1}{\sin(x)} \cdot \cos(x) \right) \cdot y \\ \dfrac{dy}{dx} \amp = \left( \dfrac{4x^3}{x^4+7} + \frac{1}{2x} - \dfrac{\cos(x)}{\sin(x)} \right) \cdot \left( \dfrac{(x^4+7)(\sqrt{x})}{\sin(x)} \right) \end{align*}