MCQ
If $y = {x^{({x^x})}}$, then ${{dy} \over {dx}} = $
- A$y[{x^x}(\log ex).\log x + {x^x}]$
- B$y[{x^x}(\log ex).\log x + x]$
- ✓$y[{x^x}(\log ex).\log x + {x^{x - 1}}]$
- D$y[{x^x}({\log _e}x).\log x + {x^{x - 1}}]$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}.\log x + \frac{1}{x}.z$ , (where ${x^x} = z$)
$ \Rightarrow \frac{{dy}}{{dx}} = {x^{({x^x})}}\left[ {{x^x}(\log ex).\log x + {x^{x - 1}}} \right]$, $\left\{ \because \frac{dz}{dx}={{x}^{x}}\log ex \right\}$.
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