MCQ
If $y = {(\sin x)^{{{(\sin x)}^{(\sin x)......\infty }}}}$, then ${{dy} \over {dx}} = $
  • ${{{y^2}\cot x} \over {1 - y\log \sin x}}$
  • B
    ${{{y^2}\cot x} \over {1 + y\log \sin x}}$
  • C
    ${{y\cot x} \over {1 - y\log \sin x}}$
  • D
    ${{y\cot x} \over {1 + y\log \sin x}}$

Answer

Correct option: A.
${{{y^2}\cot x} \over {1 - y\log \sin x}}$
a
(a) $y = {(\sin x)^{{{(\sin x)}^{(\sin x).....\infty }}}}$

==> $y = {(\sin x)^y} \Rightarrow {\log _e}y = y\log \sin x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{dy}}{{dx}}[\log \sin x + y\cot x]$

$\therefore \frac{{dy}}{{dx}} = \frac{{{y^2}\cot x}}{{1 - y\log \sin x}}$.

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