Implicit Diﬀerentiation and the Chain Rule The chain rule tells us that: d df dg (f ◦ g) = . dx dg dx While implicitly diﬀerentiating an expression like x + y 2 we use the chain rule as follows: d 2 d(y 2 ) dy (y ) = = 2yy � . dx dy dx Why can we treat y as a function of x in this way? y 2 1 0 −2 −1 0 1 2 x −1 −2 Figure 1: The hyperbola y 2 − x2 = 1. Consider the equation y 2 − x2 = 1, which describes the hyperbola shown in Figure 1. We cannot write y as a function of x, but if we start with a point (x, y) on the graph and then change its x coordinate by sliding the point along the graph its y coordinate will be constrained to change as well. The change in y is implied by the change in x and the constraint y 2 − x2 = 1. Thus, it makes dy sense to think about y � = , the rate of change of y with respect to x. dx Given that y 2 − x2 = 1: a) Use implicit diﬀerentiation to ﬁnd y � . b) Check your work by using Figure 1 to estimate the slope of the tangent line to the hyperbola when y = −1 and when x = 1. c) Check your work for y > 0 by solving for y and using the direct method to take the derivative. 1 Solution a) Use implicit diﬀerentiation to ﬁnd y � . We diﬀerentiate both sides of the equation y 2 − x2 = 1 to get: 2yy � − 2x 2yy � = = y� = y� = 0 2x 2x 2y x y b) Check your work by using Figure 1 to estimate the slope of the tangent line to the hyperbola when y = −1 and when x = 1. y 2 1 0 −2 −1 0 1 2 x −1 −2 Figure 2: Tangent lines to the hyperbola y 2 − x2 = 1. From the graph, we see that when y = −1, x = 0 and so y � = x/y = 0. This agrees with the fact that at the point (−1, 0) the tangent line to the graph is horizontal. √ √ 2. At the point (1, √2) the slope of the When x = 1, y 2 = 2 so y = ± √ tangent line is y � = x/y = 1/ 2 ≈ 2/3 and at (1, − 2) the slope of the tangent line is approximately −2/3. 2 c) Check your work for y > 0 by solving for y and using the direct method to take the derivative. We start by solving for y: y 2 − x2 = 1 2 = y = 1 + x2 � ± 1 + x2 y We’re told to consider the case y > 0, in which y = d (1 + x2 )1/2 dx = = = √ 1 + x2 = (1 + x2 )1/2 . 1 (1 + x2 )−1/2 · 2x 2 x √ 1 + x2 x y We conclude that for y > 0, diﬀerentiating directly gives the same result as implicit diﬀerentiation. 3