Question

If \( y = x^{\log x} + (\log x)^x, x>1 \) then \( (\frac{dy}{dx})_{x=e} = \)

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For functions of the form \(f(x)^{g(x)}\), always use logarithmic differentiation. Take the natural log of both sides, differentiate implicitly, and then solve for \(dy/dx\).

Answer 1

The Correct Option is D

We need to differentiate y term by term. Let \( u = x^{\log x} \) and \( v = (\log x)^x \).
For \( u = x^{\log x} \), we use logarithmic differentiation. (Assume log is natural log, ln).
\( \ln u = (\ln x)(\ln x) = (\ln x)^2 \).
Differentiating with respect to x: \( \frac{1}{u}\frac{du}{dx} = 2(\ln x) \cdot \frac{1}{x} \).
\( \frac{du}{dx} = u \cdot \frac{2\ln x}{x} = x^{\ln x} \cdot \frac{2\ln x}{x} \).
For \( v = (\ln x)^x \), we also use logarithmic differentiation.
\( \ln v = x \ln(\ln x) \).
Differentiating using the product rule: \( \frac{1}{v}\frac{dv}{dx} = (1)\ln(\ln x) + x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} \).
\( \frac{1}{v}\frac{dv}{dx} = \ln(\ln x) + \frac{1}{\ln x} \).
\( \frac{dv}{dx} = v \left( \ln(\ln x) + \frac{1}{\ln x} \right) = (\ln x)^x \left( \ln(\ln x) + \frac{1}{\ln x} \right) \).
Now, we evaluate both derivatives at \( x=e \).
\( (\frac{du}{dx})_{x=e} = e^{\ln e} \cdot \frac{2\ln e}{e} = e^1 \cdot \frac{2(1)}{e} = 2 \).
\( (\frac{dv}{dx})_{x=e} = (\ln e)^e \left( \ln(\ln e) + \frac{1}{\ln e} \right) = (1)^e \left( \ln(1) + \frac{1}{1} \right) = 1(0+1) = 1 \).
The derivative of y is the sum of the derivatives of u and v.
\( (\frac{dy}{dx})_{x=e} = (\frac{du}{dx})_{x=e} + (\frac{dv}{dx})_{x=e} = 2 + 1 = 3 \).