Answer :
Given:
[tex]x+xy=2y[/tex]You need to use Implicit Differentiation to determine the equation of a line tangent. In order to do this, you need follow these steps:
1. Derivate the function treating "y" as function of "x". Use these Derivative Rules:
- Power Rule:
[tex]\frac{d}{dx}(nx^n)=nx^{n-1}[/tex]- Product Rule:
[tex]\frac{d}{dx}(u(x)\cdot v(x))=u^{\prime}(x)^v(x)+v^{\prime}(x)u(x)[/tex]Then, you get:
[tex](x)^{\prime}+(xy)^{\prime}=(2y)^{\prime}[/tex][tex](1x^{1-1})^{\prime}+(x)^{\prime}(y)+(y)^{\prime}(x)=2y^{\prime}[/tex][tex]1+(1)(y)+xy^{\prime}=2y^{\prime}[/tex][tex]1+y+xy^{\prime}=2y^{\prime}[/tex]2. Solve for y':
[tex]1+y=2y^{\prime}-xy^{\prime}[/tex][tex]1+y=y^{\prime}(2-x)[/tex][tex]y^{\prime}=\frac{1+y}{2-x}[/tex][tex]y^{\prime}=-\frac{1+y}{x-2}[/tex]3. Having the point:
[tex](3,2)[/tex]You need to substitute its coordinate into the derivative of the function and evaluate, in order to find the slope of the line that is tangent to the given function:
[tex]y^{\prime}=-\frac{1+(2)}{(3)-2}=-\frac{3}{1}=-3[/tex]4. The Point-Slope Form of the equation of a line is:
[tex]y-y_1=m(x-x_1)[/tex]Where "m" is the slope and this is a point on the line:
[tex](x_1,y_1)[/tex]Substituting values, you get:
[tex]y-2=-3(x-3)[/tex]5. You can rewrite it in Slope-Intercept Form by solving for "y":
[tex]y=-3x+9+2[/tex][tex]y=-3x+11[/tex]Hence, the answer is:
[tex]y=-3x+11[/tex]