Implicit Diff Example - # 2

[Jason76]

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

\(\displaystyle x^{2} + 2xy - y^{2} + x = 20\)

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

\(\displaystyle 2x + 2y + 2xy' - 2yy' + 1 = 0\)

\(\displaystyle 2x + 2y + y'[2x - 2y] + 1 = 0\)

\(\displaystyle y'[2x - 2y] = -2x - 2y - 1\)

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

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

 

[stapel]

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

\(\displaystyle x^{2} + 2xy - y^{2} + x = 20\)

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

\(\displaystyle 2x + 2y + 2xy' - 2yy' + 1 = 0\)

\(\displaystyle 2x + 2y + y'[2x - 2y] + 1 = 0\)

\(\displaystyle y'[2x - 2y] = -2x - 2y - 1\)

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

It looks right to me.

What now?

Now plug in the x- and y-coordinates of the given point, and simplify to find the value of y' = dy/dx at that point. Then use what you learned back in algebra to find the straight-line equation having that slope and passing through that point.

 

[srmichael]

With implicit differentiation you always end up with both x AND y in the derivative. So, like stapel already said, you must plug in both the x AND the y value to find the slope.