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

MCQ

If $y = {x^{{x^{x......\infty }}}}$, then ${{dy} \over {dx}} = $

A
${{{y^2}} \over {x(1 + y\log x)}}$
✓
${{{y^2}} \over {x(1 - y\log x)}}$
C
${y \over {x(1 + y\log x)}}$
D
${y \over {x(1 - y\log x)}}$

✓

Answer

Correct option: B.

${{{y^2}} \over {x(1 - y\log x)}}$

b
(b) $y = {x^{{x^{x.......\infty }}}}$ ==>$y = {x^y}$==>$\log y = y\log x$

Therefore, on differentiating $\frac{{dy}}{{dx}} = \frac{{{y^2}}}{{x(1 - y\log x)}}$.

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