How do you use the chain rule to differentiate #1/ln(4x)#?

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Answer 1 · 2017-03-28T14:26:44

Let #u = 4x#, #v = ln(u) = ln(4x)#.

Differentiating the above equation yields

#frac{du}{dx} = 4#
#frac{dv}{du} = 1/u#

Then,

#1/ln(4x) = 1/v#

#frac{d}{dx}(1/ln(4x)) = frac{d}{dx}(1/v)#

#= frac{d}{dv}(1/v) frac{dv}{du} frac{du}{dx}#

#= (-1/v^2) * (1/u) * (4)#

#= (-1/(ln(4x))^2) * (1/{4x}) * (4)#

#= -1/(x (ln(4x))^2 )#