Where am I going wrong on implicit differentation?

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SUMMARY

The discussion focuses on finding the equation of a tangent line to the curve defined by the equation 3x² + xy + 2y² = 36 at the point P(2,3). The user initially calculated the derivative and obtained a slope of 1.5, but later realized the correct slope should be -15/14 after correctly substituting the point into the derivative formula. The mistake was identified as a sign error during the calculation process. The correct tangent line equation is y = -15/14x + b, where b can be determined using the point-slope form.

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Homework Statement



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

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

Homework Equations





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) into 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.
 
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Haven't looked at it carefully, but slope should be -1.5, after substitution.
 
Try replugging your point (2,3).
 
\frac{-6(2)-3}{2+4(3)}=\frac{-15}{14}

I copied the wrong sign down.
 
Last edited:
if you plug in (2,3) you get -15 / 14 for the slope.
 
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.
 

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