Question

I can not figure out how to start this problem

Answer 1

At first, we have to find y', then find y'(2)

The given equation is

`3x^2+4x+xy=2`

We want to use the rule of the product in differentiation

`uv=u^(\prime)v+uv^(\prime)`

Then to differentiate xy, it will be

`\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}`

We will differentiate it with respect to x

`3(2)x^(2-1)+4(1)x^(1-1)+(1)x^(1-1)(y)+x(1)y^(1-1)y^(\prime)=0`

Simplify each term

`\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}`

Now, we need to separate y' on the side and the other terms on the other side

`xy^(\prime)=-6x-y-4`

Divide them by x

`\begin{gathered} (xy^(\prime))/(x)=(-6x-y-4)/(x) \\ y^(\prime)=(-6x-y-4)/(x) \end{gathered}`

We need to find y'(2) by substitute x by 2 and y by -9

`\begin{gathered} y^(\prime)(2)=(-6(2)-(-9)-4)/(2) \\ y^(\prime)(2)=(-12+9-4)/(2) \\ y^(\prime)(2)=(-7)/(2) \end{gathered}`

The answer is -7/2