Asked by James
The slope of the line tangent to the curve xy+(y+1)^2=6 at the point (2, 1) is
Answers
Answered by
bobpursley
the method
x dy + y dx+2(y+1)dy = 0
dy(x+2(y+1))=-ydx
dy/dx=slope= -y/(x+2(y+1))
so figure dy/dx at the given point
check my work.
x dy + y dx+2(y+1)dy = 0
dy(x+2(y+1))=-ydx
dy/dx=slope= -y/(x+2(y+1))
so figure dy/dx at the given point
check my work.
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