Consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?
2 answers:
<span>The derivative of your function is: 1 + (1*dy/dx + y) - 4y(dy/dx) = 0
2 + dy/dx + y - 4y(dy/dx) = 0
(2 + y) + dy/dx (1 - 4y) = 0
(dy/dx) = -(2 + y)/(1 - 4y) and at point (2,1),
dy/dx = -(2 + 1)/(1 - 4*1) = -3 / -3 = 1
Slope of tangent line = 1</span>
The derivative is actually (-y-1)/(4y+x) then you plug in the points (2,1) ==> (-1-1)/(4*1+2) which equals -1/3 then you just plug in your info into a point-slope formula and get: y-1=-1/3(x-2) which can then be reduced
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