Question

find y' in x^2y+y^3+cosy=x

Answer 1

\(x^2y+y^3+\cos y=x\), find \(y'\)...

Answer 2

we simply differentiate implicitly applying rules of differentiation...
\[2xy+y'x^2+3y^2y'+(-\sin y)y'=1\]notice \(y'\) among the terms, combine like terms, re-arrange the equation...
\[(x^2+3y^2-\sin y)y'=1-2xy\]therefore
\[y'=\frac{1-2xy}{x^2+3y^2-\sin y}\]