Q2) We have to find the derivative of the function y = (ln x)^x.
We can start using the property of the derivative of an potential and exponential function:
[tex]\frac{d(u^v)}{dx}=u^v\cdot\frac{d\lbrack\ln (u)\cdot v\rbrack}{dx}[/tex]
In this case, u = ln(x) and v = x, so we can solve it as:
[tex]\begin{gathered} \frac{dy}{dx}=\ln (x)^x\cdot\frac{d\lbrack\ln (\ln (x))\cdot x\rbrack}{dx} \\ \frac{dy}{dx}=\ln (x)^x\cdot\lbrack\ln (\ln (x))\cdot1+x\frac{d(\ln (\ln (x))}{dx}\rbrack \\ \frac{dy}{dx}=\ln (x)^x\cdot\lbrack\ln (\ln (x))\cdot1+x\cdot\frac{1}{\ln(x)}\cdot\frac{1}{x}\rbrack \\ \frac{dy}{dx}=\ln (x)^x\cdot\lbrack\ln (\ln (x))+\frac{1}{\ln (x)}\rbrack \end{gathered}[/tex]
Answer: dy/dx = ln(x)^x * [ln(ln(x)) + 1/ln(x)]
