MCQ
જો $y = {\log _2}[{\log _2}(x)]$, તો ${{dy} \over {dx}}= . . . .$
- ✓${{{{\log }_2}e} \over {x{{\log }_e}x}}$
- B${1 \over {{{\log }_e}x{{\log }_e}2}}$
- C${1 \over {{{\log }_e}{{(2x)}^x}}}$
- Dએક પણ નહીં
$ = [{\log _e}{\log _e}x + {\log _e}({\log _2}e)]{\log _2}e$
$\therefore \frac{{dy}}{{dx}} = {\log _2}e.\frac{1}{{x{{\log }_e}x}}$.
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