Lesson 8: Basic Differentation Rules (handout)

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We derive some of the rules for computing derivatives.

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Lesson 8: Basic Differentation Rules (handout)

  1. 1. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Sec on 2.3 Basic Differenta on Rules V63.0121.001: Calculus I Professor Ma hew Leingang New York University . NYUMathematics . Notes Announcements Quiz 1 this week on 1.1–1.4 Quiz 2 March 3/4 on 1.5, 1.6, 2.1, 2.2, 2.3 Midterm Monday March 7 in class . . Notes Objectives Understand and use these differen a on rules: the deriva ve of a constant func on (zero); the Constant Mul ple Rule; the Sum Rule; the Difference Rule; the deriva ves of sine and cosine. . . . 1.
  2. 2. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Recall: the derivative Defini on Let f be a func on and a a point in the domain of f. If the limit f(a + h) − f(a) f(x) − f(a) f′ (a) = lim = lim h→0 h x→a x−a exists, the func on is said to be differen able at a and f′ (a) is the deriva ve of f at a. . . Notes The deriva ve … …measures the slope of the line through (a, f(a)) tangent to the curve y = f(x); …represents the instantaneous rate of change of f at a …produces the best possible linear approxima on to f near a. . . Notes Notation Newtonian nota on Leibnizian nota on dy d df f′ (x) y′ (x) y′ f(x) dx dx dx . . . 2.
  3. 3. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Link between the notations f(x + ∆x) − f(x) ∆y dy f′ (x) = lim = lim = ∆x→0 ∆x ∆x→0 ∆x dx dy Leibniz thought of as a quo ent of “infinitesimals” dx dy We think of as represen ng a limit of (finite) difference dx quo ents, not as an actual frac on itself. The nota on suggests things which are true even though they don’t follow from the nota on per se . . Notes Outline Deriva ves so far Deriva ves of power func ons by hand The Power Rule Deriva ves of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Mul ple Rule Deriva ves of sine and cosine . . Notes Derivative of the squaring function Example Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x). Solu on f(x + h) − f(x) (x + h)2 − x2 f′ (x) = lim = lim h→0 h h→0 h   + 2xh + h −   x2   2 x2   2x + h2 h ¡ = lim = lim h→0 h h→0 h = lim (2x + h) = 2x. h→0 . . . 3.
  4. 4. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes The second derivative If f is a func on, so is f′ , and we can seek its deriva ve. f′′ = (f′ )′ It measures the rate of change of the rate of change! Leibnizian nota on: d2 y d2 d2 f 2 2 f(x) dx dx dx2 . . The squaring function and its Notes derivatives y f′ f′′ f increasing =⇒ f′ ≥ 0 f decreasing =⇒ f′ ≤ 0 . f x horizontal tangent at 0 =⇒ f′ (0) = 0 . . Notes Derivative of the cubing function Example Suppose f(x) = x3 . Use the defini on of deriva ve to find f′ (x). Solu on . . . 4.
  5. 5. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . The cubing function and its Notes derivatives No ce that f is increasing, y and f′ 0 except f′′ f′ f′ (0) = 0 No ce also that the f tangent line to the graph . x of f at (0, 0) crosses the graph (contrary to a popular “defini on” of the tangent line) . . Notes Derivative of the square root Example √ Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on. Solu on √ √ f(x + h) − f(x) x+h− x f′ (x) = lim = lim h→0 √ h h→0 h √ √ √ x+h− x x+h+ x = lim · √ √ h→0 h x+h+ x (x + h) − x ¡ ¡ h 1 = lim (√ √ ) = lim (√ √ )= √ h→0 h x+h+ x h x+h+ x h→0 2 x . . The square root function and its Notes derivatives y f Here lim+ f′ (x) = ∞ and f ′ x→0 f is not differen able at 0 . x No ce also lim f′ (x) = 0 x→∞ . . . 5.
  6. 6. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Derivative of the cube root Example √ Suppose f(x) = 3 x = x1/3 . Find f′ (x). Solu on . . The cube root function and its Notes derivatives y Here lim f′ (x) = ∞ and f f x→0 is not differen able at 0 f′ . x No ce also lim f′ (x) = 0 x→±∞ . . Notes One more Example Suppose f(x) = x2/3 . Use the defini on of deriva ve to find f′ (x). Solu on . . . 6.
  7. 7. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . The function x → x2/3 and its Notes derivative y f is not differen able at 0 f and lim± f′ (x) = ±∞ f′ x→0 . x No ce also lim f′ (x) = 0 x→±∞ . . Notes Recap: The Tower of Power y y′ x2 2x1 The power goes down by x 3 3x2 one in each deriva ve 1 −1/2 The coefficient in the x1/2 2x deriva ve is the power of 1 −2/3 x1/3 3x the original func on 2 −1/3 x2/3 3x . . Notes The Power Rule There is moun ng evidence for Theorem (The Power Rule) Let r be a real number and f(x) = xr . Then f′ (x) = rxr−1 as long as the expression on the right-hand side is defined. Perhaps the most famous rule in calculus We will assume it as of today We will prove it many ways for many different r. . . . 7.
  8. 8. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Outline Deriva ves so far Deriva ves of power func ons by hand The Power Rule Deriva ves of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Mul ple Rule Deriva ves of sine and cosine . . Notes Remember your algebra Fact Let n be a posi ve whole number. Then (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) Proof. We have ∑ n (x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = ck xk hn−k n copies k=0 . . Notes Pascal’s Triangle . 1 1 1 1 2 1 1 3 3 1 (x + h)0 = 1 1 4 6 4 1 (x + h)1 = 1x + 1h (x + h)2 = 1x2 + 2xh + 1h2 1 5 10 10 5 1 (x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3 1 6 15 20 15 6 1 ... ... . . . 8.
  9. 9. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Proving the Power Rule Theorem (The Power Rule) d n Let n be a posi ve whole number. Then x = nxn−1 . dx Proof. As we showed above, (x + h)n = xn + nxn−1 h + (stuff with at least two hs in it) (x + h)n − xn nxn−1 h + (stuff with at least two hs in it) So = h h = nxn−1 + (stuff with at least one h in it) and this tends to nxn−1 as h → 0. . . Notes The Power Rule for constants? d 0 Theorem like x = 0x−1 dx d Let c be a constant. Then c = 0. . dx Proof. Let f(x) = c. Then f(x + h) − f(x) c − c = =0 h h So f′ (x) = lim 0 = 0. h→0 . . . Notes Calculus . . . 9.
  10. 10. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Recall the Limit Laws Fact Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then x→a x→a 1. lim [f(x) + g(x)] = L + M x→a 2. lim [f(x) − g(x)] = L − M x→a 3. lim [cf(x)] = cL x→a 4. . . . . . Notes Adding functions Theorem (The Sum Rule) Let f and g be func ons and define (f + g)(x) = f(x) + g(x) Then if f and g are differen able at x, then so is f + g and (f + g)′ (x) = f′ (x) + g′ (x). Succinctly, (f + g)′ = f′ + g′ . . . Notes Proof of the Sum Rule Proof. Follow your nose: (f + g)(x + h) − (f + g)(x) (f + g)′ (x) = lim h→0 h f(x + h) + g(x + h) − [f(x) + g(x)] = lim h→0 h f(x + h) − f(x) g(x + h) − g(x) = lim + lim h→0 h h→0 h = f′ (x) + g′ (x) . . . 10.
  11. 11. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Scaling functions Theorem (The Constant Mul ple Rule) Let f be a func on and c a constant. Define (cf)(x) = cf(x) Then if f is differen able at x, so is cf and (cf)′ (x) = c · f′ (x) Succinctly, (cf)′ = cf′ . . . Notes Proof of Constant Multiple Rule Proof. Again, follow your nose. (cf)(x + h) − (cf)(x) cf(x + h) − cf(x) (cf)′ (x) = lim = lim h→0 h h→0 h f(x + h) − f(x) ′ = c lim = c · f (x) h→0 h . . Notes Derivatives of polynomials Example d ( 3 ) Find 2x + x4 − 17x12 + 37 dx Solu on d ( 3 ) d ( 3) d d ( ) d 2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37) dx dx dx dx dx d d d = 2 x3 + x4 − 17 x12 + 0 dx dx dx = 2 · 3x2 + 4x3 − 17 · 12x11 = 6x2 + 4x3 − 204x11 . . . 11.
  12. 12. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Notes Outline Deriva ves so far Deriva ves of power func ons by hand The Power Rule Deriva ves of polynomials The Power Rule for whole number powers The Power Rule for constants The Sum Rule The Constant Mul ple Rule Deriva ves of sine and cosine . . Notes Derivatives of Sine and Cosine Fact d sin x = cos x dx Proof. From the defini on: d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x sin x = lim = lim dx h→0 h h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h = sin x · 0 + cos x · 1 = cos x . . Angle addition formulas Notes See Appendix A . . . . 12.
  13. 13. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Two important trigonometric Notes limits See Section 1.4 sin θ lim . =1 θ→0 θ cos θ − 1 sin θ θ lim =0 θ θ→0 θ .−1 1 − cos θ 1 . . Notes Illustration of Sine and Cosine y . x π −π 0 π π cos x 2 2 sin x f(x) = sin x has horizontal tangents where f′ = cos(x) is zero. what happens at the horizontal tangents of cos? . . Notes Derivative of Cosine Fact d cos x = − sin x dx Proof. We already did the first. The second is similar (muta s mutandis): d cos(x + h) − cos x (cos x cos h − sin x sin h) − cos x cos x = lim = lim dx h→0 h h→0 h cos h − 1 sin h = cos x · lim − sin x · lim h→0 h h→0 h = cos x · 0 − sin x · 1 = − sin x . . . 13.
  14. 14. . V63.0121.001: Calculus I . Sec on 2.3: Basic Differenta on Rules . Summary Notes What have we learned today? The Power Rule The deriva ve of a sum is the sum of the deriva ves The deriva ve of a constant mul ple of a func on is that constant mul ple of the deriva ve The deriva ve of sine is cosine The deriva ve of cosine is the opposite of sine. . . Notes . . Notes . . . 14.

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