What is the derivative of ln(lnx^2)?

1 Answer
Mar 15, 2016

d/dx [ln(lnx^2)] = 1/[xln(x)]

Explanation:

y = ln(lnx^2)

Using the chain rule:

d/dx[ln(f(x)] = [lnf(x)]^' = [1/f(x)]f'(x)

dy/dx = [1/(lnx^2)] (lnx^2)^'

= [1/(lnx^2)] (1/x^2)(x^2)^'

=[1/(lnx^2)] (1/x^2)(2x)

=2/[xln(x^2)] => simplifying:

=1/[xln(x)]