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anonymous
 5 years ago
consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?
anonymous
 5 years ago
consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11 + xy'+y+4y y' = 0 y' = 1y/(x+4y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i think your derivative is off a bit... but it could be me

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x; 1 xy; xy' + y 2y^2; 4y y' = 6; 0 1 +xy' +y +4y y' = 0 xy' +4y y' = 1y y'(x+4y) = (1y) y' = (1y)/(x+4y) ; at (x=2,y=1) y'= 1(1)/..... = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1your product rule on (xy) is a little off

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and i cant seem to add lol 11 = 2 =1/3 2+4 = 6
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