MCQ
If $y = {\sqrt x ^{{{\sqrt x }^{\sqrt x ....\infty }}}}$, then ${{dy} \over {dx}} = $
- A${{{y^2}} \over {2x - 2y\log x}}$
- B${{{y^2}} \over {2x + \log x}}$
- C${{{y^2}} \over {2x + 2y\log x}}$
- ✓None of these
==> $\log y = y\log {x^{1/2}} = \frac{1}{2}y\log x$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left( {\log x\frac{{dy}}{{dx}} + \frac{y}{x}} \right) $
$\Rightarrow \frac{{dy}}{{dx}}\left( {\frac{2}{y} - \log x} \right) = \frac{y}{x}$
==> $\frac{{dy}}{{dx}} = \frac{{y.y}}{{x(2 - y\log x)}} = \frac{{{y^2}}}{{x(2 - y\log x)}}$.
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