f(x) = x^{x} has a stationary point at | vedclass

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

Question

(B) Given the function $f(x) = x^{x}$.Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.Differentiating both sides with respect t

Vedclass · Accepted Answer

$x = \frac{1}{e}$

Vedclass · Answer

(B) Given the function $f(x) = x^{x}$.Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.Differentiating both sides with respect to $x$ using the chain rule and product rule:$\frac{1}{f(x)} f'(x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Thus,$f'(x) = f(x)(\ln x + 1) = x^{x}(\ln x + 1)$.$A$ stationary point occurs where $f'(x) = 0$.Since $x^{x} > 0$ for $x > 0$,we set $\ln x + 1 = 0$.$\ln x = -1$.Converting from logarithmic to exponential form,$x = e^{-1} = \frac{1}{e}$.

$f(x) = x^{x}$ has a stationary point at

Similar Questions

Find the derivative: $\frac{d}{dx} \log (\sqrt{x - a} + \sqrt{x - b})$

If $y = \log \left[a^{3x} \left(\frac{5-x}{x+4}\right)^{\frac{3}{4}}\right]$,then $\frac{dy}{dx} = $

If $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$,then $\frac{dy}{dx}$ is equal to

The value of $\log _{e} 2 \cdot \frac{d}{dx}(\log _{\cos x} \operatorname{cosec} x)$ at $x=\frac{\pi}{4}$ is.

$\frac{d}{dx} \left( a^{\log_{10}(\csc^{-1}x)} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module