Question

If $y=x^{\left(x^x\right)}$, then $\frac{dy}{dx}$ is

• A. $y[x^x\left(\log ex\right)\log x+x^x]$
• B. $y[x^x\left(\log ex\right)\log x+x]$
• C. $y[x^x\left(\log ex\right)\log x+x^{x-1}]$
• D. $y[x^x\left({\log }_{e}x\right)\log x+x^{x-1}]$

Answer 1

The correct option is C $y[x^x\left(\log ex\right)\log x+x^{x-1}]$

$y=x^{\left(x^x\right)}$
Take log on both sides
$\log y=\log x^{\left(x^x\right)}$
$\log y=x^x\log x$ .... $\left(i\right)$
Let $x^x=z$
$\therefore x\log x=\log z$
Differentiate both sides with respect to $x$
$\frac{1}{z}\frac{dz}{dx}=\log x+\frac{x}{x}$
$\frac{dz}{dx}=z[1+\log x]$
$\frac{dz}{dx}=x^x[\log e+\log x]$
$\frac{dz}{dx}=x^x\log ex$
Equation $\left(i\right)⟹\log y=z\log x$
Differentiate both sides with respect to $x$
$\frac{1}{y}\frac{dy}{dx}=\log x\cdot \frac{dz}{dx}+\frac{z}{x}$
$\therefore \frac{dy}{dx}=y\left[x^x\left(\log ex\right)\log x+x^{x-1}\right]$