Question

If $y=x^{x^x}$ then find,$\frac{dy}{dx}$

Answer 1

Consider $y=x^x$

$\Rightarrow \log y=\log x^x$ by applying $\log$ to both sides

$\Rightarrow \log y=x\log x$ using $\log a^m=m\log a$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\times \frac{1}{x}+\log x=1+\log x$ by differentiating both sides w.r.t $x$

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^x\right)=y(1+\log x)=x^x(1+\log x)$

Consider $y=x^{x^x}$

$\Rightarrow \log y=\log x^{x^x}$ by applying $\log$ to both sides

$\Rightarrow \log y=x^x\log x$ using $\log a^m=m\log a$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=x^x\times \frac{1}{x}+\log x\frac{d}{dx}\left(x^x\right)=x^{x-1}+\log x{x}^x(1+\log x)$ by differentiating both sides w.r.t $x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=x^{x-1}+x^x\log x(1+\log x)$

$\Rightarrow \frac{dy}{dx}=y(x^{x-1}+x^x\log x(1+\log x))$

$\Rightarrow \frac{dy}{dx}=x^{x^x}(x^{x-1}+x^x\log x+x^x{\left(\log x\right)}^2)$