I am currently learning about this very powerful calculus tool. This tool is of great value in simplifying some functions prior to differentiation. I will try to explain simply in my own words.
What are logarithms? Logarithms were invented by John Napier (1550-1617) for the purpose of simplifying calculation; basically turning a product into something simpler, such as a sum. Before the advent of calculators most calculus textbooks had a section at the back consisting of tables of logarithms. One of the most common logarithms are base 10 logarithms (\(\log_{10}{x}\), or simply \(\log {x}\)). However, for calculus, base e, also called natural, logarithms are used more frequently (\(\log_{e}{x}\), or simply \(\ln {x}\)).
What is differentiation? It is the process of obtaining the slope of the tangent line of a curve either at a specific point or obtaining a new function that ‘gives’ the slope at any point in the original function.
Then, isn’t Logarithmic Differentiation just differentiating logarithmic functions? It’s more than that. While logarithmic functions can be differentiated like any other function. Logarithmic Differentiation refers to converting almost any function into a logarithmic function prior to differentiation with the purpose of using the properties of logarithms to simplify differentiation.
What are these properties of logarithms that make differentiation simpler?
If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are positive numbers and \(\boldsymbol{n}\) is rational,
then the four properties below are true:
\begin{align}
\boxed{
\boldsymbol{1.} \ln{1} = 0 \qquad \qquad \qquad \, \; \; \boldsymbol{2.} \ln{\left(a \cdot b\right)} = \ln{a} + \ln{b} \\
\boldsymbol{3.} \ln{\left(a^{n}\right)} = n \ln{a} \qquad \qquad \boldsymbol{4.} \ln{\left(\frac{a}{b}\right)} = \ln{a}\, – \ln{b} \\
}
\end{align}
From the table above, notice that using properties 2. and 4. can turn a product or a quotient into a sum or difference. Using property 3. can turn an exponent into a coefficient.
For example, it is not an easy task to differentiate the function: \(\large{y = \frac{x^{2} \sqrt{3x\, -\, 2}}{\left(x\, +\, 1\right)^{2}} \text{, } \; x \gt \frac{2}{3}}\). To differentiate it we need to use the Product Rule, Quotient Rule, and the General Power Rule. While each of these rules is simple enough, complications arise when they are nested.
\[\text{Nevertheless, let’s give it a try: } \quad \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^{2}\, \left(3x\, -\, 2\right)^{\frac{1}{2}}}{\left(x\, +\, 1\right)^{2} } \right)\]
\begin{align*}
= \frac{\left(x\, +\, 1\right)^{2} \left(\left(x^{2} \left(\frac{1}{2}\left(3x\, -\, 2\right)^{-\frac{1}{2}}\right)\left(3\right)\right) + \left(3x\, -\, 2\right)^\frac{1}{2}\left(2x\right)\right)\, -\, \left(x^{2}\, \left(3x\, -\, 2\right)^\frac{1}{2}\right) \left(2\right)\left(x\, +\, 1\right)}{\left(x\, +\, 1\right)^{4}}
\end{align*}
We are done with the calculus part. The task now is to simplify. I am not going to show all the steps in simplifying this long statement. Anyone familiar with Algebra knows that it is not an easy task. In the end, this is what we have:
\[\frac{dy}{dx} = \frac{x \left(3x^{2} + 15x\, -\, 8\right)}{2\, \left(x + 1\right)^{3} \sqrt{3x\, -\, 2}}\]
\begin{align*}
\text{Let’s try now Algorithmic Differentiation: } \; \ln{y} &= \ln{\left(\frac{x^{2}\, \left(3x\, -\, 2\right)^{\frac{1}{2}}}{\left(x + 1\right)^{2}}\right)} \\
\text{Before differentiating, use the properties of algorithms: } \; \ln{y} &= \ln{x^{2}} + \frac{1}{2} \ln{\left(3x -\, 2\right)}\, -\, 2 \ln{\left(x + 1\right)} \\
\text{And now, let’s take the derivative on both sides: } \; \frac{\frac{dy}{dx}}{y} &= \frac{2}{x} + \frac{3}{2\, \left(3x -\, 2\right)}\, -\, \frac{2}{\left(x + 1\right)} \\
\text{Let’s place the right side under a common denominator: } \; \frac{\frac{dy}{dx}}{y} &= \frac{4\, \left(3x -\, 2\right) \left(x + 1\right) + 3x \left(x + 1\right)\, -\, 4x \left(3x -\, 2\right)}{2x \left(3x -\, 2\right) \left(x + 1\right)} \\
\text{Simplification is easier because all factors are linear: } \; \frac{\frac{dy}{dx}}{y} &= \frac{3x^{2} + 15x\, – 8}{2x \left(3x -\, 2\right) \left(x + 1\right)} \\
\text{Now. move } \boldsymbol{y} \text{ to the right side: } \; \frac{dy}{dx} &= y \cdot \left(\frac{3x^{2} + 15x\, – 8}{2x \left(3x -\, 2\right) \left(x + 1\right)}\right) \\
\text{We know what } \boldsymbol{y} \text{ is: } \; \frac{dy}{dx} &= \left(\frac{x^{2} \sqrt{3x\, -\, 2}}{\left(x\, +\, 1\right)^{2}}\right) \cdot \left(\frac{3x^{2} + 15x\, – 8}{2x \left(3x -\, 2\right) \left(x + 1\right)}\right) \\
\text{And now we have some easy cancellations: } \; \frac{dy}{dx} &= \frac{x \left(3x^{2} + 15x\, -\, 8\right)}{2\, \left(x + 1\right)^{3} \sqrt{3x\, -\, 2}}
\end{align*}
One final note: not all functions can be differentiated in this manner. The domain of logarithms is all positive numbers greater than zero. This means that \(\boldsymbol{y}\) must also be greater than zero.
Quiz #6
This week’s quiz is a ‘double-header’.
What is the domain of the following functions, their concavity, and locate the x- and y-coordinates of their minimum and/or maximum:
\[y = x \ln{x} \quad \text{ and } \quad y =\frac{\ln{x}}{x}\]
Answer to Quiz #5
Last week’s quiz was to find the exact value of \(\boldsymbol{\tan\left(75^{o}\right)}\) without using a calculator.
We had a similar quiz a few weeks ago: the trick, if there is one, is to use Trigonometric Identities; specifically, Angle Sum Identity.
\begin{align*}
\tan\left(75^{o}\right) = \tan\left(45^{o} + 30^{o}\right) &= \frac{\tan{\left(45^{o}\right)} + \tan{\left(30^{o}\right)}}{1 -\, \tan{\left(45^{o}\right)} \tan{\left(30^{o}\right)}} \\
&= \frac{1 + \frac{\sqrt{3}}{3}}{1 -\, \left(1 \cdot \left(\frac{\sqrt{3}}{3}\right)\right)} = \large{\frac{\frac{3\, +\, \sqrt{3}}{3}}{\frac{3\, -\, \sqrt{3}}{3}}} = \small{\frac{3 + \sqrt{3}}{3 -\, \sqrt{3}}} \\
\text{We could leave this here, or normalize the denominator…..} &= \frac{\left(3 + \sqrt{3}\right)}{\left(3 -\, \sqrt{3}\right)} \cdot \frac{\left(3 + \sqrt{3}\right)}{\left(3 + \sqrt{3}\right)} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}
\end{align*}
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .