Lecture 5 Sept. 15, 2006 18.01 Fall 2006 Lecture 5 : Implicit Diﬀerentiation and Inverses Implicit Diﬀerentiation d a Example 1. (x ) = axa−1 . dx We proved this by an explicit computation for a = 0, 1, 2, .... From this, we also got the formula for a = −1, −2, .... Let us try to extend this formula to cover rational numbers, as well: m a= ; n We want to compute m y=xn where m and n are integers. dy . We can say y n = xm dx so ny n−1 dy dy = mxm−1 . Solve for : dx dx dy m xm−1 = dx n y n−1 m We know that y = x( n ) is a function of x. dy dx = = = = = = = So, dy dx = � � m xm−1 n y n−1 � � m xm−1 n (xm/n )n−1 m xm−1 n xm(n−1)/n m (m−1)− m(n−1) n x n m n(m−1)−m(n−1) n x n m nm−n−nm+m n x n m m−n n xn n m m −1 xn n This is the same answer as we were hoping to get! Example√2. Equation of a circle with a radius of 1: x2 +y 2 = 1 which we can write as y 2 = 1−x2 . So y = ± 1 − x2 . Let us look at the positive case: y dy dx = = � 1 + 1 − x2 = (1 − x2 ) 2 � � −1 1 −x −x (1 − x2 ) 2 (−2x) = √ = 2 2 y 1−x 1 Lecture 5 Sept. 15, 2006 18.01 Fall 2006 Now, let’s do the same thing, using implicit diﬀerentiation. x2 + y 2 � d � 2 x + y2 dx d 2 d 2 (x ) + (y ) dx dx = 1 d (1) = 0 dx = = 0 Applying chain rule in the second term, dy = dx dy 2y = dx dy = dx Same answer! 2x + 2y 0 −2x −x y y 3 + xy 2 + 1 = 0. In this case, it’s not easy to solve for y as a function of x. Instead, dy we use implicit diﬀerentiation to ﬁnd . dx Example 3. 3y 2 We can now solve for dy dy + y 2 + 2xy = 0 dx dx dy in terms of y and x. dx dy (3y 2 + 2xy) dx dy dx = = −y 2 −y 2 + 2xy 3y 2 Inverse Functions If y = f (x) and g(y) = x, we call g the inverse function of f , f −1 : x = g(y) = f −1 (y) Now, let us use implicit diﬀerentiation to ﬁnd the derivative of the inverse function. −1 y (y) f d −1 (f (y)) dx = = = f (x) x d (x) = 1 dx By the chain rule: d −1 dy (f (y)) = dy dx and d −1 (f (y)) = dy 2 1 1 dy dx Lecture 5 Sept. 15, 2006 18.01 Fall 2006 So, implicit diﬀerentiation makes it possible to ﬁnd the derivative of the inverse function. Example. y = arctan(x) tan y d [tan(y)] dx d dy [tan(y)] dy dx � � 1 dy 2 cos (y) dx dy dx = = x dx = 1 dx = 1 = 1 = cos2 (y) = cos2 (arctan(x)) This form is messy. Let us use some geometry to simplify it. y (1+x2)1/2 1 x Figure 1: Triangle with angles and lengths corresponding to those in the example illustrating diﬀerentiation using the inverse function arctan In this triangle, tan(y) = x so arctan(x) = y The Pythagorian theorem tells us the length of the hypotenuse: � h = 1 + x2 From this, we can ﬁnd 1 cos(y) = √ 1 + x2 From this, we get 2 cos (y) = � 1 √ 1 + x2 3 �2 = 1 1 + x2 Lecture 5 Sept. 15, 2006 18.01 Fall 2006 So, dy 1 = dx 1 + x2 In other words, d 1 arctan(x) = dx 1 + x2 Graphing an Inverse Function. Suppose y = f (x) and g(y) = f −1 (y) = x. To graph g and f together we need to write g as a function of the variable x. If g(x) = y, then x = f (y), and what we have done is to trade the variables x and y. This is illustrated in Fig. 2 f −1 (f (x)) = x f (f −1 (x)) = x f −1 ◦ f (x) = x f ◦ f −1 (x) = x y f(x) y=x g(x) b=f(a) a=f-1(b) Figure 2: x You can think about f −1 as the graph of f reﬂected about the line y = x 4