Slope of a line tangent to a circle – direct version A circle of radius 1 centered at the origin consists of all points (x, y) for which x2 + y 2 = 1. This equation does not describe a function of x (i.e. it cannot be written in the form y = f (x)). Indeed, any vertical line drawn through the interior of the circle meets the circle in two points — every x has two corresponding y values. Let’s see what goes wrong if we attempt to solve the equation of a circle for y in terms of x. x2 + y 2 x2 + y 2 − x2 y2 y = 1 = 1 − x2 = 1 − x2 � = ± 1 − x2 This still isn’t a function because we get two choices for y — positive or negative. However, we do get a function if we look just at the positive case (i.e. dy at just the top half of the circle), and we can then ﬁnd dx , which will be the slope of a line tangent to the top half of the circle. To compute this derivative, we ﬁrst convert the square root into a fractional exponent so that we can use the rule from the previous example. y = � 1 1 − x2 = (1 − x2 ) 2 1 Next, we need to use the chain rule to diﬀerentiate y = (1 − x2 ) 2 . The outside function is u1/2 and the inside function is 1 − x2 , so the chain rule tells us that dy dy du = . dx du dx dy 1 −x = u−1/2 · (−2x) = −x · (1 − x2 )−1/2 = √ . dx 2 1 − x2 √ If we want, we can use the fact that y = 1 − x2 to rewrite this as y � = −x/y. We conclude that the slope of the line tangent to a point (x, y) on the top half of the unit circle is −x/y.