Find the derivative use logarithmic differentiation. \( y=x^{x} \) a) \( y^{\prime}=\ln x+1 \) b) \( y^{\prime}=x^{x}(1+\ln x) \) c) \( y \prime=\frac{(\ln x+1)}{x^{x}} \) \( { }^{\text {d) }} y \prime=x^{x}\left(\frac{1}{x}+x \ln x\right) \)

To differentiate \( y = x^x \) using logarithmic differentiation, first take the natural logarithm of both sides: \[ \ln y = \ln(x^x) = x \ln x \] Next, differentiate both sides with respect to \( x \). For the left side, apply the chain rule: \[ \frac{1}{y} \frac{dy}{dx} \] For the right side, use the product rule: \[ \frac{d}{dx}(x \ln x) = \ln x + 1 \] Equating both derivatives gives: \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \] Now, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y(\ln x + 1) = x^x (\ln x + 1) \] So the derivative is: \[ y' = x^x (1 + \ln x) \] The correct answer is option b) \( y' = x^x (1 + \ln x) \). For those who love a bit of history, did you know that logarithmic differentiation was used by mathematicians like John Napier, inventor of logarithms, who sought to simplify complex calculations in the early 17th century? Now, that's a clever twist on tackling tricky derivatives! If you're diving deeper into calculus, a fun problem-solving tip is to always remember to check if you can simplify your expressions before differentiating. Often, rewriting a complex function can uncover patterns that make taking derivatives a breeze!