Question

If $y = x^x + x^{\frac{1}{x}}$, then $\frac{dy}{dx}$ is equal to}

• A. $x^x(1 + \log x) + x^{\frac{1}{x}} \frac{1}{x^2}(1 - \log x)$
• B. $(x^x + x^{\frac{1}{x}}) \left[ 1 + \log x + \frac{1}{x^2}(1 - \log x) \right]$
• C. $(x^x + x^{\frac{1}{x}}) \left[ (1 + \log x) - \frac{1}{x^2}(1 - \log x) \right]$
• D. $x^x(1 + \log x) - x^{\frac{1}{x}} \frac{1}{x^2}(1 - \log x)$

Hint:

Memorize: $\frac{d}{dx}(x^x) = x^x(1+\ln x)$. It appears very frequently.

Answer 1

The Correct Option is A

Step 1: Concept
Use logarithmic differentiation for each term separately: $\frac{d}{dx}(u^v) = u^v \frac{d}{dx}(v \log u)$.

Step 2: Meaning
Let $u = x^x$ and $v = x^{1/x}$. $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

Step 3: Analysis
1. $\frac{du}{dx} = x^x(1 + \log x)$. 2. For $v = x^{1/x}$, $\log v = \frac{1}{x} \log x$. $\frac{1}{v} \frac{dv}{dx} = \frac{1}{x} \cdot \frac{1}{x} + \log x \cdot (-\frac{1}{x^2}) = \frac{1 - \log x}{x^2}$. $\frac{dv}{dx} = x^{1/x} \frac{1}{x^2} (1 - \log x)$.

Step 4: Conclusion
Summing the parts gives option (A). Final Answer: (A)