STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

Differentiating $\frac{{dy}}{{dx}} = {x^x}(1 + \log x)$;

$\therefore \frac{{dy}}{{dx}} = 0$

==> $\log x = - 1$==>$x = {e^{ - 1}} = \frac{1}{e}$

$\therefore $ Stationary point is $x = \frac{1}{e}$

$\frac{{{d^2}y}}{{d{x^2}}} = {x^x}{(1 + \log x)^2} + {x^x}.\frac{1}{x}$

When $x = \frac{1}{e},\;\;\frac{{{d^2}y}}{{d{x^2}}} = {\left( {\frac{1}{e}} \right)^{(1/e) - 1}} > 0$

Therefore $y$ is minimum at $x = \frac{1}{e}$ and

minimum value $ = {\left( {\frac{1}{e}} \right)^{1/e}} = {e^{ - 1/e}}$.