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Q.
If $y=e^{x^{x}}$, then $\cdot \frac{d y}{d x}=$
Continuity and Differentiability
Solution:
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
$\frac{1}{\log _{e} y} \cdot \frac{1}{y} \cdot \frac{d y}{d x}=1 \cdot \log _{e} x+x \cdot \frac{1}{x}$ or $\frac{ dy }{ dx }= y \log _{ e } y \cdot(\log x +1)$
$=e^{x^x} \cdot x^{x} .\left(\log _{e} x+1\right) .=y \cdot x^{x}\left(1+\log _{e} x\right)$