# Where am I going wrong on implicit differentation?

1. Oct 11, 2011

### jrjack

1. The problem statement, all variables and given/known data

Determine the equation of a tangent line to a curve at the given point.

3x2+xy+2y2=36 , P(2,3)

2. Relevant equations

3. The attempt at a solution

3x2+xy+2y2 = 36

finding the derivatives of each term I get:

6x+xy'+y+4yy' = 0
xy'+4yy' = -6x-y
y'(x+4y) = -6x-y
$$y' = \frac{-6x-y}{x+4y}$$
Which gives me a slope of 1.5 after plugging (2,3) in to the equation.

Then for the tangent: y-3=1.5(x-2) , y=1.5x

Somewhere I went wrong because this is not the right answer.

2. Oct 11, 2011

### Bacle

Haven't looked at it carefully, but slope should be -1.5, after substitution.

3. Oct 11, 2011

### LawrenceC

Try replugging your point (2,3).

4. Oct 11, 2011

### jrjack

$$\frac{-6(2)-3}{2+4(3)}=\frac{-15}{14}$$

I copied the wrong sign down.

Last edited: Oct 11, 2011
5. Oct 11, 2011

### dacruick

if you plug in (2,3) you get -15 / 14 for the slope.

6. Oct 11, 2011

### jrjack

Thanks for your help, I don't know why I can't seem to find my mistakes when going back through the problem. I even copied it down wrong on here as I was saying right outloud to myself.

Once again Thanks.