The maximum value of {x^{1/x}} is | vedclass

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(B) Let $y = x^{1/x}$.Taking the natural logarithm on both sides,we get $\ln y = \frac{1}{x} \ln x$.Differentiating both sides with respect to $x$ usi

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${e^{1/e}}$

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(B) Let $y = x^{1/x}$.Taking the natural logarithm on both sides,we get $\ln y = \frac{1}{x} \ln x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \left(-\frac{1}{x^2}\right) \ln x + \left(\frac{1}{x}\right) \left(\frac{1}{x}\right) = \frac{1 - \ln x}{x^2}$.Thus,$\frac{dy}{dx} = x^{1/x} \left(\frac{1 - \ln x}{x^2}\right)$.For critical points,set $\frac{dy}{dx} = 0$,which implies $1 - \ln x = 0$,so $\ln x = 1$,which gives $x = e$.Since the derivative changes from positive to negative at $x = e$,the function attains a maximum at $x = e$.The maximum value is $y(e) = e^{1/e}$.

The maximum value of ${x^{1/x}}$ is

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