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

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Answer 1 · 2014-08-21T01:52:03

Method 1:

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

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

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

The derivative of #ln x/6# will be #1/(xln6)#:

#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 #d/dx ln(x) = 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 = (log_{n}b)/(log_{n} a) = (ln b)/ln a# when #n=e#

Now, let #z = (ln x/ln 6)# so that #f(x) = z^2#

Therefore, #f'(x)=d/dx z^2 = (d/dz z^2)( dz/dx) = 2z d/dx (ln x/ln 6)#

#= (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)#