xy^2 = 1
x(2y)dy/dx + y^2 (1) = 0
dy/dx = -y^2/(2xy)
at (1,-1) , dy/dx = -(-1)^2/(2(1)(-1))= 1/-2 = - 1/2
Now you have the slope and the given point. Find the
equation of the straight line, using your method of choice.
Find the equations of the tangent lines to the following curves at the indicated points.
xy^2 = 1 at (1, −1)
1 answer
1 answer