The problem is f(x)=(ln x^2)^2
I looked at the answer key and the answer is 4(ln x)^2/x
I do not get how the numerator can be squared.
I only got 4(ln x).
and how the denominator becomes x, instead of 2x?
f(x)=(ln x^2)^2 = (2lnx)^2 = 4 (lnx)^2
f'(x) = 4*2(lnx)*(1/x) = 4ln(x^2)/x
Take it as a compound function, like:
g(u(x))=u(x)^2 where u(x)=ln x^2.
In that case:
f'(x) = 2u*u'(x) = 2u*(1/x^2)*(x^2)'=(2u/x^2)*2x
Either your answer key is wrong OR You are typing it wrongly here. OR I may be wrong
Thank you for the correction. The result should remain the same though.
One more detail. Answer key: 4(ln x)^2/x is definitely wrong! Cause it is not defined for negative numbers. However f(x) is defined everywhere except zero, and differentiable everywhere except zero.