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QuarkCharmer

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## Homework Statement

Stewart Calculus (6E) 3.6 question #26

Using implicit differentiation, find the equation of a tangent line to the curve at the given point.

[tex]x^2+2xy-y^2+x=2[/tex]

Through point p(1,2)

I just want to know if I am doing this correctly.

## Homework Equations

## The Attempt at a Solution

I start with the original equation, and find the derivative with respect to x. (I am assuming x right?)

[tex]x^2+2xy-y^2+x=2[/tex]

[tex]\frac{d}{dx}x^2+2xy-y^2+x=2[/tex]

[tex]2x+2(y+xy')-2yy'+1=0[/tex]

[tex]2x+2y+2xy'-2yy'+1=0[/tex]

Now I isolate the derivative of y(y') and solve for it:

[tex]2x+2y+1=y'(2y-2x)[/tex]

[tex]y'=\frac{2x+2y+1}{2y-2x}[/tex]

So, to get the derivative (slope of tangent) at that point, I put that x/y coordinate in for the derivative equation and get y' :

[tex]\frac{2(1)+2(2)+1}{2(2)-2(1)} = \frac{2+4+1}{4-2} = \frac{7}{2}[/tex]

Now, having the slope of the tangent line at that point on the original curve, I can just stick it in the equation of a line with the coordinate points.

[tex]2=\frac{7}{2}(1)+k[/tex]

[tex]2-\frac{7}{2} = k[/tex]

[tex]k = -\frac{3}{2}[/tex]

So the solution is:

[tex]y=\frac{7}{2}x-\frac{3}{2}[/tex]

?

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