# Thread: Implicit Diff Example - # 2

1. ## Implicit Diff Example - # 2

Find the equation of the tangent line at the point (3,4)

$x^{2} + 2xy - y^{2} + x = 20$

$2x + [(y)(2) + (2x)(y')] - 2yy' + 1 = 0$

$2x + 2y + 2xy' - 2yy' + 1 = 0$

$2x + 2y + y'[2x - 2y] + 1 = 0$

$y'[2x - 2y] = -2x - 2y - 1$

$y' = \dfrac{-2x - 2y - 1}{2x - 2y}$ - Did this come out right?

What now? We have two variables so how to find the slope?

2. ## Re: Implicit Diff Example - # 2

Hey Jason76.

Looks very good: now just plug in the values for x and y to get the tangent gradient and recall that y - y0 = m(x - x0) for specific (x0,y0) on the line and general (x,y) on the line.

3. ## Re: Implicit Diff Example - # 2

Find the equation of the tangent line at the point (3,4)

$x^{2} + 2xy - y^{2} + x = 20$

$2x + [(y)(2) + (2x)(y')] - 2yy' + 1 = 0$

$2x + 2y + 2xy' - 2yy' + 1 = 0$

$2x + 2y + y'[2x - 2y] + 1 = 0$

$y'[2x - 2y] = -2x - 2y - 1$

$y' = \dfrac{-2x - 2y - 1}{2x - 2y}$

$y' = \dfrac{-2(3) - 2(4) - 1}{2(3) - 2(4)}$

$y' = \dfrac{-6 - 8 - 1}{6 - 8}$

$y' = \dfrac{-15}{-2}$

$y' = \dfrac{15}{2}$

$y - 4 = \dfrac{15}{2}(x - 3)$

$y - 4 = \dfrac{15}{2}x -\dfrac{45}{2}$ ??