Thread: Derivatives of log function!

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?

Thanks lots!

f(x)=(ln x^2)^2 = (2lnx)^2 = 4 (lnx)^2
f'(x) = 4*2(lnx)*(1/x) = 4ln(x^2)/x

OR

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
=4u/x
=4(lnx^2)/x

Either your answer key is wrong OR You are typing it wrongly here. OR I may be wrong

-O
4(ln x)^2/x

Originally Posted by oswaldo

f(x)=(ln x^2)^2 = (2lnx)^2 = 4 (lnx)^2
[snip]

. .

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.

-O