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!

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- Feb 19th 2009, 11:31 PMkatieeejDerivatives 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! - Feb 19th 2009, 11:47 PMoswaldo
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 - Feb 20th 2009, 04:07 AMmr fantastic
- Feb 20th 2009, 06:55 AMoswaldo
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