Q1757: Find the values of x for which $f(x)=x^{x}$, $x>0$ is increa...

CBSE Class 12 Mathematics Applications of Derivatives Q #1757

COMPETENCY BASED

APPLY

2 Marks 2026 AISSCE(Board Exam) VSA

Find the values of x for which $f(x)=x^{x}$, $x>0$ is increasing.

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Detailed Solution

Step 1: Logarithmic Differentiation

Given the function $f(x) = x^x$. To differentiate this, we take the natural logarithm on both sides: $$\ln(f(x)) = \ln(x^x) = x \ln(x)$$

Step 2: Differentiate with respect to x

Differentiating both sides with respect to $x$: $$\frac{1}{f(x)} \cdot f'(x) = \frac{d}{dx}(x \ln x)$$ Using the product rule: $$\frac{f'(x)}{f(x)} = (1 \cdot \ln x) + (x \cdot \frac{1}{x}) = \ln x + 1$$ Thus, $f'(x) = x^x(1 + \ln x)$.

Step 3: Determine the condition for increasing

A function is increasing when $f'(x) > 0$. Since $x^x > 0$ for all $x > 0$, the sign of $f'(x)$ depends solely on $(1 + \ln x)$: $$1 + \ln x > 0$$ $$\ln x > -1$$ $$x > e^{-1} = \frac{1}{e}$$

Final Answer: x > 1/e

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student must apply the rules of logarithmic differentiation and the first derivative test for monotonicity.

Knowledge Dimension: PROCEDURAL

Justification: The question requires a specific sequence of mathematical steps (logarithmic transformation, differentiation, and inequality solving) to reach the solution.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to handle transcendental functions within the Application of Derivatives unit.