Question

The function $x^x$ is increasing, when

• A. $x>\frac{1}{e}$
• B. $x<\frac{1}{e}$
• C. $x<0$
• D. for all $x$

Hint:

For $x^x$, derivative is $x^x(\ln x + 1)$—remember this shortcut.

Answer 1

The Correct Option is A

Concept: Use logarithmic differentiation for $x^x$.

Step 1: Take logarithm.
\[ y = x^x \Rightarrow \ln y = x\ln x \]

Step 2: Differentiate.
\[ \frac{1}{y}\frac{dy}{dx} = \ln x + 1 \] \[ \frac{dy}{dx} = x^x(\ln x + 1) \]

Step 3: Check increasing condition.
\[ \frac{dy}{dx}>0 \Rightarrow \ln x + 1>0 \] \[ \ln x>-1 \Rightarrow x>\frac{1}{e} \] Conclusion:
Function increases when $x>\frac{1}{e}$