differentiation

Question: If $y = {\sqrt x ^{{{\sqrt x }^{\sqrt x ....\infty }}}}$, then $\frac{{dy}}{{dx}} = $



1) $\frac{{{y^2}}}{{2x - 2y\log x}}$
2) $\frac{{{y^2}}}{{2x + \log x}}$
3) $\frac{{{y^2}}}{{2x + 2y\log x}}$
4) None of these
Solution: Explanation: Explanation
Implicit-function

Rate this question:

Average rating: (0 votes)

Previous Question Next Question