\(y=x^{x^2}\)
\(\ln y=\ln x^{x^2}\)
\(\ln y=x^2\ln x\)
Now differentiating with respect to \(x\) gives
\(\dfrac{y'}y=2x\ln x+\dfrac{x^2}x\)
\(\dfrac{y'}y=2x\ln x+x\)
\(\dfrac{y'}y=x(\ln x^2+1)\)
\(y'=yx(\ln x^2+1)\)
\(y'=x^{x^2}x(\ln x^2+1)\)
\(y'=x^{x^2+1}(\ln x^2+1)\)
