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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

${e^{1/e}}$

Vedclass · Answer

(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

Similar Questions

Let $A$ be the region enclosed by the parabola $y^2=2x$ and the line $x=24$. Then the maximum area of the rectangle inscribed in the region $A$ is ..................

The number of real roots of the equation $e^{4x} + 2e^{3x} - e^{x} - 6 = 0$ is:

Let $f(x) = \int\limits_0^x \frac{\cos t}{t} dt, x > 0$. Then $f(x)$ has:

If $y = \frac{ax - b}{(x - 1)(x - 4)}$ has a turning point $P(2, -1)$,then the values of $a$ and $b$ are

The function $f(x) = 2x + 3x^{\frac{2}{3}}, x \in R$,has

Vedclass Test Series

Exam Paper Generator

Online Exam Module