Answer
$y^{\prime}=\displaystyle \frac{xy^{2}+y}{x\ln x}$
Work Step by Step
Differentiating an equation containing
terms of form $f(x)$ and $g(y)$,
terms $f(x)$ are differentiated directly, $\displaystyle \frac{d}{dx}[f(x)],$
and
terms $g(y)$ are differentiated using the chain rule,
$\displaystyle \frac{d}{dx}[g(y)]=\frac{d}{dy}[g(y)]\cdot\frac{dy}{dx}$
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LHS: quotient, $( \displaystyle \frac{u}{v})^{\prime}=\frac{u^{\prime}v-uv^{\prime}}{v^{2}}$
RHS: sum, first term:constant
$\displaystyle \frac{\ln x}{y}=2-x$
Differentiate both sides,
$\displaystyle \frac{\frac{1}{x}\cdot y-\ln x\cdot y^{\prime}}{y^{2}}=0-1\qquad/\times y^{2}$
... isolating $y^{\prime}$ on one side...
$\displaystyle \frac{y}{x}-\ln x\cdot y^{\prime}=-y^{2}\qquad/-\frac{y}{x}$
$-\displaystyle \ln x\cdot y^{\prime}=-y^{2}-\frac{y}{x}\qquad/\times(-1)$
$\displaystyle \ln x\cdot y^{\prime}=y^{2}+\frac{y}{x}\qquad/\div\ln x$
$y^{\prime}=\displaystyle \frac{xy^{2}+y}{x\ln x}$
