We have,
2x+3y=sinx
On differentiating and we get,
ddx(2x+3y)=ddxsinx
⇒2dxdx+3dydx=cosx
⇒3dydx=cosx−2
⇒dydx=cosx−23
Hence, this is the answer.
Find dydxin the following questions:
2x+3y=sin x.