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Sagot :
Given :
A curve : 3x² - 2xy + y² = 11
To Find :
The slope of the line tangent to the curve 3x^2-2xy+y^2=11 at the point (1, -2).
Solution :
Differentiating given curve w.r.t x is :
6x - ( 2y + 2xy' ) + 2yy' = 11
Putting the point ( 1, -2) in above equation, we get :
6 - ( 2(-2) + 2y' ) - 4y' = 11
6 - ( -4 + 2y' ) - 4y' = 11
10 - 6y' = 11
6y' = -1
[tex]y' =\dfrac{-1}{6}[/tex]
Therefore, the slope of the line tangent to the given point is [tex]y' =\dfrac{-1}{6}[/tex] .
Answer:
5/3
Step-by-step explanation:
The previous explanation was incorrect because they forgot to differentiate the other side of the equation (11), which would actually become 0, and make the answer 5/3
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