18.01 Calculus Jason Starr Fall 2005 Lecture 4. September 15, 2005 Homework. No new problems. Practice Problems. Course Reader: 1F1, 1F6, 1F7, 1F8. √ 1. Product rule example. For u = 3x + 1, what is u� (x)? Since u · u = 3x + 1, (u · u)� = (3x + 1)� = 3. By the product rule, (u · u)� = u� · u + u · u� = 2uu� . Thus solving, u� (x) = 3/(2u) = 3(3x + 1)−1/2 /2 . 2. The derivative of un . From above, (u2 )� equals 2uu� . By a similar computation, (u3 )� equals 3u2 u� . This suggests a pattern, d(un ) du = nun−1 . dx dx 18.01 Calculus Jason Starr Fall 2005 This can be proved by induction on n. For n = 1, 2 and 3, it was checked. Let n be a particular integer (for instance, 70119209472933054321). For that integer, suppose the result is known, d(un ) du = nun−1 . dx dx The goal is to prove the result for n + 1, that is, d(un+1 ) du = (n + 1)un . dx dx Let v = un . Then un+1 equals uv. So, by the product rule, d(un+1 ) d(uv) du dv = = v+u . dx dx dx dx Plugging in v = un , this is, d(un+1 ) du d(un ) = . · (un ) + u dx dx dx By the induction hypothesis, d(un )/dx equals nun−1 (du/dx). Plugging in, d(un+1 ) du du = · (un ) + u(nun−1 ). dx dx dx This simplﬁes to, d(un+1 ) du du du = un + nun = (n + 1)un . dx dx dx dx Thus, the result for n + 1 follows from the result for n. By induction, the result holds for every n. 3. The derivative of xa , a a fraction. Let a be a fraction m/n and let u(x) be xa . Then un equals xm . Thus, d(un ) d(xm ) = , dx dx which equals mxm−1 . By the above, d(un )/dx equals nun−1 (du/dx). Thus, nun−1 du = mxm−1 . dx Solving for du/dx, mxm−1 du mxm−1 = = . dx nun−1 n(xm/n )n−1 One of the basic rules of exponents is that (ab )c equals abc . Thus the denominator n(xm/n )n−1 equals nxm/n(n−1) , which equals nxm−m/n . Thus, m m−1 m/n−m du mxm−1 ·x = = x . dx nxm−m/n n 18.01 Calculus Jason Starr Fall 2005 Another basic rule of exponents is that ab · ac equals ab+c . Thus, du m m = x(m−1)+(m/n−m) = xm/n−1 . dx n n Remembering that m/n is just a, and u(x) is xa , this ﬁnally gives, d(xa ) = axa−1 . dx 4. The chain rule. Let y be a function of x, y = f (x), and let u be a function of y, u = g(y). Then u is a function of x, u = g(f (x)). This function is a composite function, and is denoted by, (g ◦ f )(x) = g(f (x)). What is the derivative of a composite function? The claim is that, (g ◦ f )� (x) = g � (f (x)) · f � (x). This is often easier to remember in the form, du du dy = · . dx dy dx This also suggests the proof, Δu Δu Δy = lim · , Δx→0 Δx Δx→0 Δy Δx (g ◦ f )� (x0 ) = lim where y0 equals f (x0 ), u0 equals g(y0 ) = g(f (x0 )), Δy equals f (x0 + Δx) − f (x0 ) = f (x0 + Δx) − y0 , and Δu equals g(y0 +Δy)−g(y0 ) = g(f (x0 +Δx))−g(f (x0 )). So long as Δy is nonzero, the fraction in the limit is deﬁned. And, as Δx approaches 0, also Δy approaches 0. Thus the limit breaks up as, Δu Δy = g � (y0 ) · f � (x0 ). · lim (g ◦ f )� (x0 ) = lim Δy→0 Δy Δx→0 Δx Thus (g ◦ f )� (x0 ) equals g � (f (x0 ))f � (x0 ). Example. Let y(x) equals 1 + x2 , and let u(y) equal 1/y = y −1 . Then y � (x) = 0 + 2x = 2x and u� (y) = −y −2 . Thus, by the chain rule, � � −1 d 1 −2x = (2x) = (1+x 2 )2 . dx 1 + x2 y2 5. Implicit diﬀerentiation. This method has already been used many times. Given a function y(x) satisfying some equation involving both x and y, formally diﬀerentiate each side of the equation with respect to x and then try to solve for y � .