Question

If \( y = x^x + x^x \), then find \( \frac{dy}{dx} \):

• A. \( x^x(\ln x + 1) \)
• B. \( 2x^x(\ln x + 1) \)
• C. \( x^x(\ln x - 1) \)
• D. \( 2x^x \ln x \)

Hint:

To differentiate \( x^x \), use logarithmic differentiation. Always remember:
If \( y = x^x \), then \( \frac{dy}{dx} = x^x(\ln x + 1) \).

Answer 1

The Correct Option is B

Provided: \[ y = x^x + x^x = 2x^x \] To find the derivative of \( x^x \), we employ logarithmic differentiation: Let \( u = x^x \). Taking the natural logarithm of both sides yields \( \ln u = x \ln x \). Differentiating implicitly with respect to \( x \): \[ \frac{1}{u} \cdot \frac{du}{dx} = \ln x + 1 \Rightarrow \frac{du}{dx} = x^x(\ln x + 1) \] Given \( y = 2x^x \), its derivative is: \[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(x^x) = 2x^x(\ln x + 1) \]