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# 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.