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MIT 18 01 - Chain Rule, and Higher Derivatives

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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 4 Sept. 14, 2006 18.01 Fall 2006 Lecture 4 Chain Rule, and Higher Derivatives Chain Rule We’ve got general procedures for differentiating expressions with addition, subtraction, and multi- plication. What about composition? Example 1. y = f(x) = sin x, x = g(t) = t2 . So, y = f(g(t)) = sin(t2). To find dy , write dt t = t0 + Δtt0 = t0 x = x0 + Δxx0 = g(t0) y = y0 + Δyy0 = f(x0) Δy =Δy Δx Δt Δx · Δt As Δt 0, Δx 0 too, because of continuity. So we get: → → dy dy dx = The Chain Rule! dt dx dt ← In the example, dx dt = 2t and dy dx = cos x. So, d dt � sin(t2) � = ( dy dx )( dx dt ) = = (cos x)(2t) (2t) � cos(t2) � Another notation for the chain rule � � d dt f(g(t)) = f�(g(t))g�(t) or d dx f(g(x)) = f�(g(x))g�(x) Example 1. (continued) Composition of functions f(x) = sin x and g(x) = x2 (f g)(x) = f(g(x)) = sin(x 2)◦ (g f)(x) = g(f (x)) = sin2(x)◦ Note: f ◦ g �= g ◦ f. Not Commutative! 1� � � � � � Lecture 4 Sept. 14, 2006 18.01 Fall 2006 xgg(x)f(g(x))fFigure 1: Composition of functions: f g(x) = f(g(x))◦ d 1Example 2. cos = ? dx x 1Let u = x dy = dy du dx du dx dy du 1 du = − sin(u); dx = − x2 � �1 � � sin dy sin(u) x = = (− sin u) −1= dx x2 x2 x2 d � � Example 3. x−n = ? dx � �n1 1There are two ways to proceed. x−n = , or x−n = x xn 1. d � x−n � = d � 1 �n = n � 1 �n−1 � −1 � = −nx−(n−1)x−2 = −nx−n−1 dx dx x x x2 2. d � x−n � = d 1= nx n−1 −1= −nx−n−1 (Think of xn as u)dx dx xn x2n 2� � Lecture 4 Sept. 14, 2006 18.01 Fall 2006 Higher Derivatives Higher derivatives are derivatives of derivatives. For instance, if g = f�, then h = g� is the second derivative of f. We write h = (f�)� = f��. Notations f�(x) f��(x) f���(x) f(n)(x) Df D2f D3f Dnf df dx d2f dx2 d3f dx3 dnf dxn Higher derivatives are pretty straightforward —- just keep taking the derivative! nExample. Dnx = ? Start small and look for a pattern. Dx = 1 D2 x 2 = D(2x) = 2 ( = 1 2) · D3 x 3 = D2(3x 2) = D(6x) = 6 ( = 1 2 3)· · D4 x 4 = D3(4x 3) = D2(12x 2) = D(24x) = 24 ( = 1 2 3 4)· · · Dn x n = n! we guess, based on the pattern we’re seeing here. ← The notation n! is called “n factorial” and defined by n! = n(n − 1) 2 1· · · · Proof by Induction: We’ve already checked the base case (n = 1). nInduction step: Suppose we know Dnx = n! (nth case). Show it holds for the (n + 1)st case. Dn+1 x n+1 = Dn Dxn+1= Dn ((n + 1)x n) = (n + 1)Dn x n = (n + 1)(n!) Dn+1 x n+1 = (n + 1)! Proved!


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MIT 18 01 - Chain Rule, and Higher Derivatives

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