The derivative of y = x^{left(x^xright)} with | vedclass

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(C) Let $y = x^{\left(x^x\right)}$. Taking the natural logarithm on both sides,we get: $\log y = x^x \log x$. Now,differentiate both sides with respec

Vedclass · Accepted Answer

$x^{\left(x^x\right)} \cdot x^x \left( \frac{1}{x} + \log x (1 + \log x) \right)$

Vedclass · Answer

(C) Let $y = x^{\left(x^x\right)}$. Taking the natural logarithm on both sides,we get: $\log y = x^x \log x$. Now,differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (x^x) \cdot \log x + x^x \cdot \frac{d}{dx} (\log x)$. To find $\frac{d}{dx} (x^x)$,let $u = x^x$. Then $\log u = x \log x$. Differentiating gives $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. So,$\frac{du}{dx} = x^x (1 + \log x)$. Substituting this back into the main equation: $\frac{1}{y} \frac{dy}{dx} = [x^x (1 + \log x)] \log x + x^x \cdot \frac{1}{x}$. $\frac{dy}{dx} = y \left[ x^x \log x (1 + \log x) + x^x \cdot \frac{1}{x} \right]$. $\frac{dy}{dx} = x^{\left(x^x\right)} \cdot x^x \left( \frac{1}{x} + \log x (1 + \log x) \right)$.

The derivative of $y = x^{\left(x^x\right)}$ with respect to $x$ is:

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