How do you find the derivative of #y=ln ln(2x^4)#?

1 Answer
Andrea S.
Dec 4, 2016

#(dy)/(dx) = 4/(xln(2x^4))#

Explanation:

Using the chain rule we know that in general:

#d/(dx) ln f(x) = (f'(x))/f(x)#

Apply this formula iteratively:

#(dy)/(dx) = (d(lnln(2x^4)))/(dx) = 1/ln(2x^4) (d(ln(2x^4)))/(dx) = #

# = 1/ln(2x^4)1/(2x^4)(d(2x^4))/(dx) = 1/ln(2x^4)*1/(2x^4)*8x^3 = 4/(xln(2x^4))#

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