# Logarithmic Differentiation - when to use?

Sorry if this is an ignorant or uninformed question, but I would like to know when I can (or should use) logarithmic differentiation. I haven't taken calculus in a while so I'm quite rusty.

So, let's say I have the following equation:

$$y = \sqrt[3]{x^2 +3} / \sqrt[]{x^2 +1}$$

I know how to solve this using logarithmic differentiation, but I'm also wondering if it'd be acceptable, or plausible, to solve using the quotient rule.

Similarly, for equations that I can solve using various rules (like chain rule, product rule, etc), am I also allowed to used logarithmic differentiation instead? My understanding of it remains vague.

Thanks!

• If your expression involved say $4$ multiplications/divisions, logarithmic differentiation would be the way to go. With $2$, the need is much less. Sep 24, 2014 at 18:37

With differentiation, you have many techniques. You are free to use whichever technique you are most comfortable with to find the derivative because all techniques give the same result.

It would be acceptable to find the derivative in any of the ways you mentioned. You could even use the product rule or the limit definition if you so choose.

That problem is one where logarithmic differentiation is especially helpful but it will never be necessary unless you are specifically asked to use logarithmic differentiation in the context of a test or homework.

Likewise, you can always use the technique of logarithmic differentiation to solve a problem but it might not be of very much use in all cases.

• Great, thank you! That was the answer I was hoping for.
– Ryan
Sep 24, 2014 at 18:35

A nice (and popular) application of a logarithmic derivative $(\log(f)'=\frac{f'}{f}$ is the formula $$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{n\ge 1}\Lambda(n)n^{-s}$$ for the Riemann Zeta function, with $\Re(s)\ge 1$, and the von Mangoldt function $\Lambda$. For a proof see here.

A typical function for using logarithmic differentiation is $f(x)=x^x$. When I was young I reinvented that method just to be able to derive that function.