where u=8x
Thus,
You might want to try it again with that.
The reason that is the same as
is because due to the properties of logs. So ln(8x) is a constant added to ln(x)
Here is what wolfram alpha says:
http://www.wolframalpha.com/input/?i...al+2%2F(xln(8x))
and the attached pdf file is what I did.
What I do not get is why you don't divide it by 8 like I multiplied by x for the du/dx algebraic manipulation.
Any help would be greatly appreciated!
Thanks in advance!