If y = x ^ x ^ x .... , then dy dx = - Vidyadip

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

✓

Answer

Correct option: D.

None of these

d
(d) $y = {\sqrt x ^{{{\sqrt x }^{\sqrt x .....\infty }}}} \Rightarrow y = {(\sqrt x )^y}$

==> $\log y = y\log {x^{1/2}} = \frac{1}{2}y\log x$

==> $\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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