Lecture 3 Notes: Differentiation Formulas: Products and Quotients

18.01 Calculus Jason Starr Fall 2005 Lecture 3. September 13, 2005 Homework. Problem Set 1 Part I: (i) and (j). Practice Problems. Course Reader: 1E­1, 1E­3, 1E­5. √ 1. Another derivative. Use the 3­step method to compute the derivative of f (x) = 1/ 3x + 1 is, f � (x) = −3(3x + 1)−3/2 /2 . Upshot: Computing derivatives by the definition is too much work to be practical. We need general methods to simplify computations. 18.01 Calculus Jason Starr Fall 2005 2. The binomial theorem. For a positive integer n, the factorial, n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1, is the number of ways of arranging n distinct objects in a line. For two positive integers n and k, the binomial coefficient, � � n n! n(n − 1) · · · (n − k + 2)(n − k + 1) = = , k k!(n − k)! k(k − 1) · · · 3 · 2 · 1 is the number of ways to choose a subset of k elements from a collection of n elements. A funda­ mental fact about binomial coefficients is the following, � � � � � � n n n+1 + = . k−1 k k This is known as Pascal’s formula. This link is to a webpage produced by MathWorld, part of Wolfram Research. The Binomial Theorem says that for every positive integer n and every pair of numbers a and b, (a + b)n equals, � � n n−k k n n−1 a + na b + · · · + a b + · · · + nabn−1 + bn . k This is proved by mathematical induction. First, the result is very easy when n = 1; it just says that (a + b)1 equals a1 + b1 . Next, make the induction hypothesis that the theorem is true for the integer n. The goal is to deduce the theorem for n + 1, � � n + 1 n+1−k k n+1 n+1 n a b + · · · + (n + 1)abn + bn+1 . (a + b) =a + (n + 1)a b + · · · + k By the definition of the (n + 1)st power of a number, (a + b)n+1 = (a + b) × (a + b)n . By the induction hypothesis, the second factor can be replaced, � � � � n n−k k n n n (a + b)(a + b) = (a + b) a + · · · + a b + ··· + b . k Multiplying each term in the second factor first by a and then by b gives, � � � n � n−k k+1 an+1 + nan b + . . . + � nk �an+1−k bk + �k+1 + ... � a b n n n−k k+1 n n+1−k k + a b + . . . + k−1 a b + a b + ... k + abn + nabn + bn+1 Summing in columns gives, an+1 + (n + 1)an b + . . . � � � n � n+1−k k � n � �n� n−k k+1 + ( nk + k−1 )a b + ( k+1 + k )a b + ... + (1 + n)abn 18.01 Calculus Jason Starr Fall 2005 Using Pascal’s formula, this simplifies to, � � n+1−k k � � n−k k+1 a b + n+1 a b + ... an+1 + (n + 1)an b + . . . + n+1 k+1 k + (n + 1)abn + bn+1 . This proves the theorem for n + 1, assuming the theorem for n. Since we proved the theorem for n = 1, and since we also proved that for each integer n, the theorem for n implies the theorem for n + 1, the theorem holds for every integer. 3. The derivative of xn . Let f (x) = xn where n is a positive integer. For every a and every h, the binomial theorem gives, � � n n−k k n n n−1 f (a + h) = (a + h) = a + na h + · · · + a h + · · · + hn . k Thus, f (a + h) − f (a) equals, n n (a + h) − a = na n−1 � � n n−k k h + ··· + a h + · · · + hn . k Thus the difference quotient is, � � � � f (a + h) − f (a) n n−2 n n−k k−1 n−1 = na + a h + ··· + a h + · · · + hn−1 . h 2 k Every summand except the first is divisible by h. The limit of such a term as h → 0 is 0. Thus, f (a + h) − f (a) = nan−1 + 0 + · · · + 0 = nan−1 . h→0 h lim So f � (x) equals nxn−1 . 3. Linearity. For differentiable functions f (x) and g(x) and for constants b and c, bf (x) + cg(x) is differentiable and, (bf (x) + cg(x))� = bf � (x) + cg � (x). This is often called linearity of the derivative. 4. The Leibniz rule/Product rule. For differentiable functions f (x) and g(x), the product f (x)g(x) is differentiable and, (f (x)g(x))� = f � (x)g(x) + f (x)g � (x). The crucial observation in proving this is rewriting the increment of f (x)g(x) from a to a + h as, f (a+h)g(a+h)−f (a)g(a) = f (a+h)[g(a+h)−g(a)]+f (a+h)g(a)−f (a)g(a) = f (a+h)[g(a+h)−g(a)]+[f (a+h)− 5. The quotient rule. Let f (x) and g(x) be differentiable functions. If g(a) is nonzero, the quotient function f (x)/g(x) is defined and differentiable at a, and, (f (x)/g(x))� = [f � (x)g(x) − f (x)g � (x)]/g(x)2 . 18.01 Calculus Jason Starr Fall 2005 One way to deduce this formula is to set q(x) = f (x)/g(x) so that f (x) = q(x)g(x), and the apply the Leibniz formula to get, f � (x) = q � (x)g(x) + q(x)g � (x) = q � (x)g(x) + f (x)g � (x)/g(x). Solving for q � (x) gives, q � (x) = [f � (x) − f (x)g � (x)/g(x)]/g(x) = [f � (x)g(x) − f (x)g � (x)]/g(x)2 . 6. Another proof that d(xn )/dx equals nxn−1 . This was mentioned only very briefly. The product rule also gives another induction proof that for every positive integer n, d(xn )/dx equals nxn−1 . For n = 1, we proved this by hand. Let n be some specific positive integer, and make the induction hypothesis that d(xn )/dx equals nxn−1 . The goal is to deduce the formula for n + 1, d(xn+1 ) = (n + 1)xn . dx By the Leibniz rule, d(xn+1 ) d(x × xn ) d(x) n d(xn ) d(xn ) = = x +x = (1)xn + x . dx dx dx dx dx By the induction hypothesis, the second term can be replaced, d(xn+1 ) = xn + x(nxn−1 ) = xn + nxn = (n + 1)xn . dx Thus the formula for n implies the formula for n + 1. Therefore, by mathematical induction, the formula holds for every positive integer n.