Use logarithms for variable powers

$\ln y=(\sin x)\ln x$. Thus $y^{\prime}/y=\sin x/x+(\ln x)\cos x$. Multiplying by $y=x^{\sin x}$ gives $y^{\prime}=x^{\sin x}[\sin x/x+(\ln x)\cos x]$, retaining all printed steps and the positive-domain condition.

The Student copy leaves the working blank. The Teacher takes logarithms: $y\ln x=x\ln y$. Differentiating both products gives $y/x+(\ln x)y^{\prime}=(x/y)y^{\prime}+\ln y$.

$(\ln x-x/y)y^{\prime}=\ln y-y/x$. Equivalently, $[(y\ln x-x)/y]y^{\prime}=(x\ln y-y)/x$. Hence the printed result is $y^{\prime}=y(x\ln y-y)/[x(y\ln x-x)]$, on a differentiable local branch where $y\ln x-x\ne0$.

The displayed quotient does not determine a slope when $y\ln x-x=0$. For example, at $(e,e)$ both its numerator and denominator vanish. The diagonal branch $y=x$ satisfies the relation and has slope 1 there; the quotient alone cannot resolve the singular point. This qualification is independently authored; the original Teacher formula remains preserved.