## Calculus (3rd Edition)

$$h'=t^tt^{t^t}\left(\frac{1}{t}+(1+\ln t )\ln t\right).$$
To find the derivative, we have $$\ln h= \ln t^{t^t}=t^t\ln t\Longrightarrow \frac{h'}{h}=t^t \frac{1}{t}+(t^t)'\ln t.$$ Now, to find the derivative of $g=t^t$, we get $$\ln g=t\ln t\Longrightarrow \frac{g'}{g}=\ln t + 1\Longrightarrow g'=t^t(1+\ln t ).$$ Finally, we have $$h'=t^{t^t}\left(t^t \frac{1}{t}+t^t(1+\ln t )\ln t\right)=t^tt^{t^t}\left(\frac{1}{t}+(1+\ln t )\ln t\right).$$