If $y=\sin(x^x)$ for $x>0$, what is $\dfrac{dy}{dx}$?

Underlying principle: this requires both logarithmic differentiation and the chain rule. 1. Let $u=x^x$. Then $y=\sin u$, so $\dfrac{dy}{dx}=\cos u\cdot \dfrac{du}{dx}$. 2. To find $\dfrac{du}{dx}$, write $u=x^x$. Taking logs, $\log u=x\log x$. 3. Differentiate: $\dfrac{1}{u}\dfrac{du}{dx}=1+\log x$, hence $\dfrac{du}{dx}=x^x(1+\log x)$. 4. Substitute back: $\dfrac{dy}{dx}=\cos(x^x)\cdot x^x(1+\log x)$. Why others are wrong: A misses the factor $(1+\log x)$ from differentiating $x^x$. B uses $\sin$ instead of $\cos$ for the derivative of the outer function. D omits the factor $x^x$.