If y = x^{(x^x)},then frac{dy}{dx} = | vedclass

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(C) Given $y = x^{(x^x)}$.Taking natural logarithm on both sides: $\log y = x^x \log x$.Differentiating both sides with respect to $x$:$\frac{1}{y} \f

Vedclass · Accepted Answer

$y[x^x(\log_e x + 1)\log x + x^{x-1}]$

Vedclass · Answer

(C) Given $y = x^{(x^x)}$.Taking natural logarithm on both sides: $\log y = x^x \log x$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x) \cdot \log x + x^x \cdot \frac{d}{dx}(\log x)$.We know that if $z = x^x$,then $\log z = x \log x$,so $\frac{1}{z} \frac{dz}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.Thus,$\frac{dz}{dx} = x^x(\log x + 1)$.Substituting this back: $\frac{1}{y} \frac{dy}{dx} = [x^x(\log x + 1)] \log x + x^x \cdot \frac{1}{x}$.Since $x^x \cdot \frac{1}{x} = x^{x-1}$,we have:$\frac{dy}{dx} = y [x^x(\log x + 1)\log x + x^{x-1}]$.

If $y = x^{(x^x)}$,then $\frac{dy}{dx} = $

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