Question

If \( y = (\log x)^{1/x} + x^{\log x} \), then at \( x = e \), \( \frac{dy}{dx} \) equals:

• A. \( 2 + \frac{1}{e} \)
• B. \( e^2 + \frac{1}{2} \)
• C. \( \frac{1}{e^2} + 2 \)
• D. \( \frac{1}{2e} + 2 \)

Hint:

When differentiating composite expressions involving logs and exponents, take the natural logarithm and apply chain rule carefully. Evaluate at the required point only after simplifying derivatives.

Answer 1

The Correct Option is C

Step 1: Differentiate the function term by term.
Let \[ y = (\log x)^{1/x} + x^{\log x} \] Step 2: Differentiate \( f(x) = (\log x)^{1/x} \).
Let \( f(x) = u(x)^{v(x)} = (\log x)^{1/x} \) Take logarithm: \[ \ln f = \frac{1}{x} \cdot \ln(\log x) \Rightarrow \frac{1}{f} f'(x) = -\frac{1}{x^2} \ln(\log x) + \frac{1}{x \log x} \cdot \frac{1}{x} \] Now: \[ f'(x) = (\log x)^{1/x} \left( -\frac{1}{x^2} \ln(\log x) + \frac{1}{x^2 \log x} \right) \] At \( x = e \): \[ \log e = 1, \quad \ln(\log e) = \ln 1 = 0 \Rightarrow f'(e) = (1) \cdot \left( 0 + \frac{1}{e^2 \cdot 1} \right) = \frac{1}{e^2} \] Step 3: Differentiate \( g(x) = x^{\log x} \).
\[ x^{\log x} = e^{\log x \cdot \ln x} \Rightarrow \frac{d}{dx} = e^{(\ln x)^2} \cdot \frac{2 \ln x}{x} \Rightarrow \text{At } x = e, \quad g'(e) = e^{1^2} \cdot \frac{2}{e} = e \cdot \frac{2}{e} = 2 \] Step 4: Add both results:
\[ \frac{dy}{dx} = f'(e) + g'(e) = \frac{1}{e^2} + 2 \]