![2x+3y=sinx\\\\3y=sinx-2x\ \ \ \ /:3\\\\y=\frac{1}{3}(sinx-2x)\\\\\frac{\delta y}{\delta x}=\left[\frac{1}{3}(sinx-2x)\right]'=\frac{1}{3}\left[(sinx)'-(2x)'\right]=\frac{1}{3}(cosx-2)](/tpl/images/0748/2213/5eb11.png)
Find dy/dx
2x+3y=sinx...
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