How do you differentiate #y= ln (ln6x)#?

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Answer 1 · 2016-04-02T20:28:10

#dy/dx = 1/(xln(6x))#

You will have to use the chain rule twice, as there is one embedded function inside another: #6x# inside #ln# inside another #ln#.

The chain rule states that

#f(x) = g(h(x))#
#f'(x) = h'(x)g'(h)#

Let's solve for the internal #ln(6x)# first.

#h'(x) = d/dx 6x = 6#
#g'(h) = 1/(h(x)) = 1/(6x)#
#d/dx ln(6x) = 6/(6x) = 1/x#

That is just the inside function. The derivative of the whole thing will need to use the chain rule again.

#f'(x) = h'(x)g'(h)#
#h'(x) = 1/x#
#g'(h(x)) = 1/(h(x)) = 1/ln(6x)#

#h'(x)g'(h) = 1/x * 1/ln(6x) = 1/(xln(6x))#