At first, we have to find y', then find y'(2)
The given equation is
[tex]3x^2+4x+xy=2[/tex]We want to use the rule of the product in differentiation
[tex]uv=u^{\prime}v+uv^{\prime}[/tex]Then to differentiate xy, it will be
[tex]\begin{gathered} u=x,v=y \\ xy\rightarrow(1)x^{1-1}(y)+x(1)y^{1-1}y^{\prime} \\ xy\rightarrow x^0y+xy^0y^{\prime} \\ xy\rightarrow(1)y+x(1)y^{\prime} \\ xy\rightarrow y+xy^{\prime} \end{gathered}[/tex]We will differentiate it with respect to x
[tex]3(2)x^{2-1}+4(1)x^{1-1}+(1)x^{1-1}(y)+x(1)y^{1-1}y^{\prime}=0[/tex]Simplify each term
[tex]\begin{gathered} 6x^1+4x^0+x^0y+xy^0y^{\prime}=0 \\ 6x+4(1)+(1)y+x(1)y^{\prime}=0 \\ 6x+4+y+xy^{\prime}=0 \end{gathered}[/tex]Now, we need to separate y' on the side and the other terms on the other side
[tex]xy^{\prime}=-6x-y-4[/tex]Divide them by x
[tex]\begin{gathered} \frac{xy^{\prime}}{x}=\frac{-6x-y-4}{x} \\ y^{\prime}=\frac{-6x-y-4}{x} \end{gathered}[/tex]We need to find y'(2) by substitute x by 2 and y by -9
[tex]\begin{gathered} y^{\prime}(2)=\frac{-6(2)-(-9)-4}{2} \\ y^{\prime}(2)=\frac{-12+9-4}{2} \\ y^{\prime}(2)=\frac{-7}{2} \end{gathered}[/tex]The answer is -7/2
