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

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Answer 1 · 2016-12-04T01:34:05

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

Using the chain rule we know that in general:

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

Apply this formula iteratively:

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

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

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