MCQ
${d \over {dx}}({x^{{{\log }_e}x}}) = $
- ✓$2{x^{({{\log }_e}x - 1)}}.{\log _e}x$
- B${x^{({{\log }_e}x - 1)}}$
- C${2 \over x}{\log _e}x$
- D${x^{({{\log }_e}x - 1)}}.{\log _e}x$
==> ${\log _e}y = {\log _e}x{\log _e}x = {({\log _e}x)^2}$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = 2{\log _e}x.\frac{1}{x}$
$\therefore \frac{{dy}}{{dx}} = 2{x^{({{\log }_e}x - 1)}}{\log _e}x$.
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Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.
If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1{ }^2+\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to....................