Derivative rules vs algebra rules for natural log

Question

I have come across a confusing situation while learning the derivative rules. Natural logarithm has two rules which are confusing to me:

• $D[\ln u] = \frac{1}{u} \cdot D[u]$ derivative rule.
• $\ln[a^b] = b \cdot \ln a$ algebra rule.

I have a hard time distinguishing which rule to use during calculations. Two questions in my mind:

1. Is $\ln(3x^2)$ a $\ln u$ (natural log of function, $u$) or not (using natural log's power rule)?
2. What about $\ln((2x + 3)^2)$? which rule to apply here?

Am I supposed to de-compose further by using rules or am I able to get the derivative now?

It does get more complicated when radicals are involved. Square radicals are re-written as a power of half. Other types of expressions that can be re-written as a base to a power also raise these questions.

I have asked myself

• Is it about the parenthesis? This sounds silly because I believe $\ln(3x^2) \equiv \ln3x^2$.
• Perhaps they are the same? A dangerous assumption during simplifications which yields different results.
• Maybe it is about the power? Does the power belong to the whole or part of a polynomial? I don't know if this is the distinguishing factor. It might be.

Background

I was well on my way to finish my Calculus course, then I arrived at the concept of derivatives. In a sense it seems more complicated than integration, but I could be wrong. I really hope derivatives are the most complicated topic to master in my Calculus course.

Answer 1

There's a lot of good points to bring up here. Let's begin with some basic "operator precedence" and how parentheses and other grouping symbols help us.

When we write $3x^2$, we mean that we square the value of $x$ then multiply that result by $3.$ Now compare to $(3x)^2,$ where we multiply by $3$ first, then square. Now, if we want to take the derivative of the function $f(x) = \ln (3x^2)$ (I put parentheses here to indicate the argument of the logarithm is $3x^2$ for clarity), we can just use the chain rule. That is,

$$ f'(x) = \frac{6x}{3x^2} = \frac{2}{x} $$

or, if we wanted to, we could also do

$$ f'(x) = \frac{d}{dx} \left( \ln 3 + \ln (x^2) \right)$$

and when you finish it, you get the same result.

For your second point, I would prefer to write $\ln \left( (2x+3)^2 \right)$ (note the extra set of parentheses). This is just for clarity's sake. Again, we use the chain rule, or we can be lazy and kick the exponent out first, then take derivatives. You'll get the same result.

Logarithms are nice tools - using their properties can help clean up some nasty calculations. In fact, you can avoid the quotient rule in some cases if you take logarithms first and use implicit differentiation!