5. Find the slope of the line which is tangent to the curve consisting of points in the plane satisfying the following equation
Find the slope of the line which is tangent to the curve at the following points.
a) 
(1, 2) 
b) 
(1/2, 1) 
To find dy/dx, differentiate both sides of the equation with respect to x.
Transpose the terms that do not invove the unknown dy/dx to the other side of the equation
Factor out the unknown dy/dx from the terms on the left
Divide both sides by the coefficient of the unknown.
a) At (1, 2)
In this case, since the equation is a quadratic in y, it is relatively easy to solve for y as a function of x.
Transpose all terms to one side
and use the quadratic formula
Incredibly, the square root comes out even.
which gives us two answers
Indeed the equation factors
When we set the factors equal to 0, we get,




If we draw the graph, we see that (1, 2) is on the portion of the graph that is given by
and that the slope of the tangent line at that point is 1.
If we differentiate y = (x + 1)/x, we get, using the chain rule
and we see that when x = 1, the slope of the tangent line is 1.
b) Also looking at the graph, we see that the point (1/2, 1) is on both parts of the graph. The question becomes, for which part of the graph will the formula for dy/dx give us the slope of the tangent line. If you look at our formula for
if we substitute in the coordinates of (1/2, 1) we get
which is what we often get when the answer can be any one of several possibilities.