What is the derivative of #log_2(3x-1)#?

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Answer 1 · 2015-09-10T04:57:13

#d/dx log_2(3x-1) = 3/(3xln 2-ln2) #

You might already know that:

#d/dx ln x = 1/x#

In order to use this, we have to apply the change of base rule:

#log_a(b) = log_c(b)/log_c(a)#

Therefore:

#log_2(3x-1)#

#= ln (3x-1)/ ln 2#

Now we can differentiate the expression:

#d/dx ln (3x-1)/ ln 2 #

#1/ln 2 # is a constant, so we can bring it out:

#d/dx ln (3x-1)/ ln 2 #
#= 1/ln 2 * d/dx ln (3x-1) #

Next, we can use the chain rule:

#(f @ g)'(x) = f'(g(x)) * g'(x)#

1) Take the derivative of the outer function..
2) ...plug in the inner function.
3) Multiply by the derivative of the inner function.

So:

#1/ln 2 * d/dx ln (3x-1) #

#= 1/ln 2 * 1/(3x-1) * d/dx(3x-1) #

#= 1/ln 2 * 1/(3x-1) * (3)#

#= 3/((ln 2)(3x-1)) #

#= 3/(3xln 2-ln2) #