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

##### 1 Answer

**Method 1:**

We will begin by using the change-of-base rule to rewrite

#f(x) = (lnx/ln6)^2#

We know that

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

So, we will apply the chain rule:

#f'(x) = 2*(lnx/ln6)^1 * d/dx[ln x / ln 6]#

The derivative of

#f'(x) = 2*(lnx/ln6)^1 * 1 / (xln 6)#

Simplifying gives us:

#f'(x) = (2lnx)/(x(ln6)^2)#

**Method 2:**

The first thing to note is that **only**

We must therefore convert the

#log_a b = (log_{n}b)/(log_{n} a) = (ln b)/ln a# when#n=e#

Now, let

Therefore,

#= (2z)/(ln 6) d/dx ln x = (2z)/(ln 6) 1/x#

#= (2/ln 6)(ln x/ln 6)(1/x) = (2 ln x)/(x*(ln 6)^2)#