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 . dy , write So, y = f (g(t)) = sin(t2 ). To ﬁnd dt t0 = t0 x0 = g(t0 ) y0 = f (x0 ) t = t0 + Δt x = x0 + Δx y = y0 + Δy 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 dy = 2t and = cos x. dt dx So, � d � sin(t2 ) dt dy dx )( ) dx dt (cos x)(2t) � � (2t) cos(t2 ) = ( = = Another notation for the chain rule d f (g(t)) = f � (g(t))g � (t) dt Example 1. (continued) � or � d � � f (g(x)) = f (g(x))g (x) dx Composition of functions f (x) = sin x and g(x) = x2 (f ◦ g)(x) (g ◦ f )(x) Note: f ◦ g = = �= f (g(x)) g(f (x)) g ◦ f. 1 = sin(x2 ) = sin2 (x) Not Commutative! Lecture 4 Sept. 14, 2006 x g Figure 1: Example 2. Let u = d cos dx g(x) 18.01 Fall 2006 f f(g(x)) Composition of functions: f ◦ g(x) = f (g(x)) � � 1 =? x 1 x dy dx dy du dy dx Example 3. = dy du du dx = − sin(u); = du 1 = − 2 dx x � � 1 � � sin sin(u) −1 x = (− sin u) = x2 x2 x2 d � −n � =? x dx There are two ways to proceed. x−n = � �n 1 1 , or x−n = n x x � �n � �n−1 � � 1 1 −1 = n = −nx−(n−1) x−2 = −nx−n−1 x x x2 � � � � d � −n � d 1 −1 n−1 2. x = = nx = −nx−n−1 (Think of xn as u) dx dx xn x2n d � −n � d 1. x = dx dx 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) Df f �� (x) D2 f f �� (x) D3 f f (n) (x) Dn f df dx d2 f dx2 d3 f dx3 dn f dxn Higher derivatives are pretty straightforward —- just keep taking the derivative! Example. Dn xn = ? Start small and look for a pattern. Dx D x D 3 x3 D 4 x4 D n xn 2 2 = = = = = 1 D(2x) = 2 ( = 1 · 2) D2 (3x2 ) = D(6x) = 6 ( = 1 · 2 · 3) D3 (4x3 ) = D2 (12x2 ) = D(24x) = 24 ( = 1 · 2 · 3 · 4) 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). Induction step: Suppose we know Dn xn = n! (nth case). Show it holds for the (n + 1)st case. � � Dn+1 xn+1 = Dn Dxn+1 = Dn ((n + 1)xn ) = (n + 1)Dn xn = (n + 1)(n!) Dn+1 xn+1 = (n + 1)! Proved! 3