MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � Lecture 5 18.01 Fall 2006 Lecture 5 Implicit Differentiation and Inverses Implicit Differentiation dExample 1. (x a) = ax a−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 m a = ; y = xn where m and n are integers. n We want to compute dy . We can say yn = xm so nyn−1 dy = mx m−1 . Solve for dy : dx dx dx dy = m xm−1 dx n yn−1 ( m We know that y = xn ) is a function of x. dy = m xm−1 dx n yn−1 m xm−1 = n (xm/n)n−1 m xm−1 = nxm(n−1)/n = x(m−1)− m(nn −1)m n m n(m−1)−m(n−1) = x n n m nm−n−nm+m = x n n m m n = x n − n n dy m m So, = x n − 1 dx n This is the same answer as we were hoping to get! Example 2. Equation of a circle with a radius of 1: x2 +y2 = 1 which we can write as y2 = 1−x2 . So y = ±√1 − x2. Let us look at the positive case: � 1 y = + 1 − x2 = (1 − x 2) 2 dy = 1(1 − x 2) −21 (−2x) = −x = −x dx 2 √1 − x2 y 1Lecture 5 18.01 Fall 2006 Now, let’s do the same thing, using implicit differentiation. x 2 + y 2 = 1 d � 2� d x 2 + y = (1) = 0 dx dx d d(x 2) + (y 2) = 0 dx dx Applying chain rule in the second term, 2x + 2ydy = 0 dx 2ydy = −2x dx dy = −x dx y Same answer! Example 3. y3 + xy2 + 1 = 0. In this case, it’s not easy to solve for y as a function of x. Instead, we use implicit differentiation to find dy . dx 3y 2 dy + y 2 + 2xydy = 0 dx dx We can now solve for dy in terms of y and x. dx dy dx (3y 2 + 2xy) = −y 2 dy = −y2 dx 3y2 + 2xy 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. y = f(x) f−1(y) = x d d(f−1(y)) = (x) = 1 dx dxBy the chain rule: d dy(f−1(y)) = 1 dy dx and d 1 (f−1(y)) = dy dy dx 2� Lecture 5 18.01 Fall 2006 So, implicit differentiation makes it possible to find the derivative of the inverse function. Example. y = arctan(x) tan y = x d dx [tan(y)] = dx dx = 1 d dy [tan(y)] � 1 cos2(y) � dy dx dy dx = = 1 1 dy dx = cos2(y) = cos2(arctan(x)) This form is messy. Let us use some geometry to simplify it. 1x(1+x2)1/2yFigure 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 1cos(y) = √1 + x2 From this, we get � �21 1 cos2(y) = =√1 + x2 1 + x2 3Lecture 5 18.01 Fall 2006 So, dy = 1 dx 1 + x2 In other words, d 1arctan(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−1 f(x) = x◦ f(f−1(x)) = x f f−1(x) = x◦ f(x)g(x)a=f-1(b)b=f(a)xyy=xFigure 2: You can think about f −1 as the graph of f reflected about the line y = x
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