# What is the derivative of f(x)=(log_6(x))^2 ?

Aug 21, 2014

Method 1:

We will begin by using the change-of-base rule to rewrite $f \left(x\right)$ equivalently as:

$f \left(x\right) = {\left(\ln \frac{x}{\ln} 6\right)}^{2}$

We know that $\frac{d}{\mathrm{dx}} \left[\ln x\right] = \frac{1}{x}$.

(if this identity looks unfamiliar, check some of the videos on this page for further explanation)

So, we will apply the chain rule:

$f ' \left(x\right) = 2 \cdot {\left(\ln \frac{x}{\ln} 6\right)}^{1} \cdot \frac{d}{\mathrm{dx}} \left[\ln \frac{x}{\ln} 6\right]$

The derivative of $\ln \frac{x}{6}$ will be $\frac{1}{x \ln 6}$:

$f ' \left(x\right) = 2 \cdot {\left(\ln \frac{x}{\ln} 6\right)}^{1} \cdot \frac{1}{x \ln 6}$

Simplifying gives us:

$f ' \left(x\right) = \frac{2 \ln x}{x {\left(\ln 6\right)}^{2}}$

Method 2:

The first thing to note is that only $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$ where $\ln = {\log}_{e}$. In other words, only if the base is $e$.

We must therefore convert the ${\log}_{6}$ to an expression having only ${\log}_{e} = \ln$. This we do using the fact

${\log}_{a} b = \frac{{\log}_{n} b}{{\log}_{n} a} = \frac{\ln b}{\ln} a$ when $n = e$

Now, let $z = \left(\ln \frac{x}{\ln} 6\right)$ so that $f \left(x\right) = {z}^{2}$

Therefore, $f ' \left(x\right) = \frac{d}{\mathrm{dx}} {z}^{2} = \left(\frac{d}{\mathrm{dz}} {z}^{2}\right) \left(\frac{\mathrm{dz}}{\mathrm{dx}}\right) = 2 z \frac{d}{\mathrm{dx}} \left(\ln \frac{x}{\ln} 6\right)$

$= \frac{2 z}{\ln 6} \frac{d}{\mathrm{dx}} \ln x = \frac{2 z}{\ln 6} \frac{1}{x}$
$= \left(\frac{2}{\ln} 6\right) \left(\ln \frac{x}{\ln} 6\right) \left(\frac{1}{x}\right) = \frac{2 \ln x}{x \cdot {\left(\ln 6\right)}^{2}}$