Question
यदि $y = {x^{({x^x})}}$, तो $\frac{{dy}}{{dx}} = $
==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}.\log x + \frac{1}{x}.z$ , (जहाँ ${x^x} = z$)
$ \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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