Question
यदि $y = {x^x}$, तो $\frac{{dy}}{{dx}} = $
दोनों पक्षों का $\log $ लेने पर, $\log y = x\log x$
अब $x$ के सापेक्ष अवकलन करने पर,
$\frac{1}{y}\frac{{dy}}{{dx}} = 1 + \log x$;
$\therefore \frac{{dy}}{{dx}} = {x^x}(1 + \log x) = {x^x}\log ex$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.