In summary, to differentiate (x^2)^{ln x}, we can use the chain rule to write it as e^{2(ln x)^2} and then differentiate to get (4ln x)/x * e^{2(ln x)^2} or we can take the logarithm of both sides and differentiate to get (4ln x)/x * (x^2)^{ln x}.
Yes that's correct. You can write the part in the exponent slightly easier by using [itex]\ln x^2=2\ln x[/itex]. Now you just have to use the chain rule.
I 'said' that, because in the latex code I wrote 2\lnx, \lnx is not a command in latex so it doesn't recognize it. I added the space now so it becomes 2\ln x, which should display the correct result.
Edit: I see you edited post 3, but what you did there is not correct. [tex](x^2)^{\ln x}=(x)^{2 \ln x}=(e^{\ln x})^{\ln x^2}=e^{\ln x \ln x^2} \neq e^{(\ln x^2)^{\ln x}}[/tex].
Another way to differentiate [itex]y= (x^2)^{ln x}[/itex] is to take the logarithm of both sides: [itex]ln(y)= ln((x^2)^{ln(x)}= ln(x)(ln(x)^2)= 2 (ln(x))^2[/itex]
Now, differentiating both sides with respect to x,
[tex]\frac{1}{y}y'= 4 ln(x)\frac{1}{x}[/tex]
[tex]y'= \frac{4 ln(x)}{x} y= \frac{4 ln(x)}{x}(x^2)^{ln(x)}[/tex]
(If you were given the problem as [itex](x^2)^{ln x}[/itex] it is probably better to write the answer using that rather than [itex]e^{2(ln x)^2}[/itex].)