Question

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

• A. $y\left(1+{\log }_{e}x\right)$
• B. $y{x}^x\left(1+{\log }_{e}x\right)$
• C. $y{e}^x\left(1+{\log }_{e}x\right)$
• D. None of these

Answer 1

Answer: (B)

Now, $y=e^{x^x}$
Taking logarithms with base e, we get
${\log }_{e}y={\log }_e{e}^{x^x}$
${\log }_{e}y=x^x\cdot {\log }_{e}e=x^x,\left\{\because {\log }_{e}e=1\right\}.$
Again taking logarithms with base e, we get,
${\log }_e\left({\log }_{e}y\right)={\log }_e{x}^x$ or ${\log }_e\left({\log }_{e}y\right)=x{\log }_{e}x$
Differentiating both sides with respect to $x$, we get
$\frac{1}{{\log }_{e}y}\cdot \frac{1}{y}\cdot \frac{dy}{dx}=1\cdot {\log }_{e}x+x\cdot \frac{1}{x}$ or $\frac{dy}{dx}=y{\log }_{e}y\cdot (\log x+1)$
$=e^{x^x}\cdot x^x.\left({\log }_{e}x+1\right).=y\cdot x^x\left(1+{\log }_{e}x\right)$