Note04 - Outline Notation of the derivative...

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Unformatted text preview: Outline Notation of the derivative Differentiability and continuity Lecture 4 - Differentiation AMA140/150 Calculus Basic rules of differentiation Derivative of inverse function Logarithmic function and exponential function Differentiation of parametric function Implicit differentiation AMA140/150 Calculus Lecture 4 - Differentiation Notation of the derivative AMA140/150 Calculus Lecture 4 - Differentiation Notation of the derivative Notation of the derivative Let f be a function defined on an interval I . We say that f is differentiable at a point x0 in I if the limit lim f x0 x →0 x − f x0 x exists. The limit, when it exists, is called the derivative of f at x0 , and is denoted by f ′ x0 . The expression at x0 . f x0 x −f x0 x , where AMA140/150 Calculus x 0, is called a difference quotient of f Lecture 4 - Differentiation Geometrical interpretation: the derivative f ′ x0 , if exists, is the slope of the tangent line to the group of the function y f x at the point x0 , f x0 . Physical interpretation: Suppose an object moves along a straight line and its distance from a fixed point on the line at time t is given by y f t . Then the derivative f ′ t0 is the instantaneous velocity of the object at the time t0 . AMA140/150 Calculus Lecture 4 - Differentiation Notation of the derivative Suppose y dy dx Derivative f x . It is also common to use the notation df dx x x0 f ′ x0 x0 x − f x0 x x →0 x x0 Notice that we put x f x0 lim lim x →x0 f x − f x0 . x − x0 Example 4.2 Let y d3 x dx dy dx x in the last equality. x 3 . Then for any x , fx x →0 Derivative function f′ x lim Example 4.1 For the constant function f x f′ x lim fx x −f x x x →0 ′ Thus, f x x →0 lim 0 x →0 ′ AMA140/150 Calculus 3x 0 and Lecture 4 - Differentiation AMA140/150 Calculus Solution: We have to find f ′ 0 x −f 0 x lim |x |? f0 x −f 0 . Notice that x x →0 lim Example 4.4 Let f x 0 x −0 x x →0 lim 1 x →0 lim f0 x −f 0 x Therefore, lim f0 lim x →0 − −0 x −0 x 1, lim −1 x → 0− x −f 0 does not exist and hence f is not x x →0 differentiable at 0. if x ≥ 0, x2 x x f′ 2 2 3x 2 . 12. if x < 0. Solution. Again we have to find f ′ 0 lim f0 x −f 0 x x →0 x → 0− 3 Differentiability Example 4.3 Is f differentiable at 0 if f x x →0 x Lecture 4 - Differentiation x6 f0 2 x − x3 0. Differentiability lim 3 In particular, f′ 0 C x 3x 2 for all x . Thus, f ′ x 0. 2 x x lim 3x 2 0 for all x . Equivalently, d C dx 3x x →0 x →0 C −C x 3x x 3x 2 x lim C, lim 3x 2 x x →0 x −f x . x x →0 x3 lim Given a function f . Suppose D is the set of all points at which f is differentiable. The derivative function of f , f ′ , with domain D is defined by fx x 3 − x3 x x lim −1. lim f0 lim lim lim f0 x →0 x −f 0 . Notice that x x 6 − 06 x 0 x →0− x →0 x 2 − 02 x 0 x →0 x −f 0 x x → 0− Is f differentiable at 0? x → 0− x →0 Remark that x if x > 0, fx −x if x < 0. ⇒ 1 if x > 0, −1 ′ if x < 0. In this case, fx f′ x 5 6x if x > 0, 0 2x AMA140/150 Calculus Lecture 4 - Differentiation x lim x x −f 0 exists and equals to 0. Hence f is x 0. differentiable at 0 and f ′ 0 Therefore, lim f0 lim AMA140/150 Calculus if x 0, if x < 0. Lecture 4 - Differentiation 5 0, 0. Differentiability and Continuity Basic rules of differentiation Differentiability ⇒ Continuity Basic rules of differentiation If f is differentiable at x0 , then f is continuous at x0 . Suppose f and g are differentiable at a point x and c is a constant. Then f x − f x0 x 0, we have ′ cf′ x ; f x − f x0 · x − x0 . x − x0 Then lim f x − f x0 · x − x0 x − x0 lim lim f x − f x0 x →x0 f x − f x0 x − x0 x →x0 x →x0 f ′ x0 · 0 Hence, lim f x x →x0 · lim x − x0 1 cf 2 f g ′ x f′ x 3 Proof. Suppose f is differentiable at x0 . For any f −g ′ x f ′ x − g′ x ; 4 fg 5 x →x0 ′ f g x g′ x ; f′ x g x x ′ f x g′ x (Product Rule); f ′ x g x − f x g′ x g2 x x (Quotient Rule). 0. f x0 . Thus, f is continuous at x0 . AMA140/150 Calculus Lecture 4 - Differentiation Derivatives of some common functions Derivatives of some common functions d 1 C 0 for any constant C ; dx dr 2 x r x r −1 for any nonzero real number r ; dx d 3 sin x cos x ; dx d 4 cos x − sin x ; dx d 5 tan x sec2 x ; dx d 6 cot x − cot2 x ; dx d 7 sec x sec x tan x ; dx d 8 csc x − csc x cot x . dx AMA140/150 Calculus Lecture 4 - Differentiation AMA140/150 Calculus Lecture 4 - Differentiation Derivatives of some common functions Example 4.5 Find the derivatives of the following functions. √ 1 3 x− x 1 fx 5x 2 2 gx 3 sin x x3 4 x 3 sin x hx kx x 4 sin x cos x 10x 3 √ 2x Solution. 1 f′ x 2 g′ x 3 h′ x 4 k′ x 5 2x 3 1 1 −2 x 2 3x 2 sin x − −x −2 x 3 cos x . cos x x 3 − sin x 3x 2 x6 x cos x − 3 sin x . x4 4x 3 sin x cos x 4x 3 sin x cos x 1 . x2 x 3 cos x − 3x 2 sin x x6 x 4 cos x cos x x 4 sin x − sin x x 4 cos2 x − x 4 sin2 x . AMA140/150 Calculus Lecture 4 - Differentiation Chain Rule Chain Rule Chain Rule Suppose h is differentiable at x0 and g is differentiable at h x0 . Let f Then f is differentiable at x0 and ′ ′ f x0 g h x0 g ◦ h. g hx x →x0 x →x0 lim x →x0 g h x0 h x − h x0 · x − x0 · lim x →x0 h x − h x0 x − x0 cos3 x kx cos sin x 2 x Solution. 2 f′ x 3x g′ x cos 2x · 3 h′ x 3 cos x 4 k′ x sin sin x 2 1 2 · d x dx 1 d 2x dx · 1 2 2 cos 2x d cos x dx · 3x −3 cos2 x sin x 1 d sin x 2 dx sin sin x 2 1 cos x 2 · −2x sin sin x 2 AMA140/150 Calculus d2 x dx cos x 2 Lecture 4 - Differentiation Derivative of inverse function Example 4.7 Find the derivatives of the following functions. ′ Suppose f is differentiable at x0 with f x0 y0 f x0 and f −1 ′ y0 0. Then f −1 is differentiable at 1 1 . f ′ x0 d sin−1 y dy 2 d cos−1 y dy Solution. Let f x sin x . Then f −1 y derivative of inverse function, Sketch of proof. Notice that f −1 ◦ f x d sin−1 y dy x. Differentiating both sides of the equation and applying Chain rule, d f −1 ◦ f x dx x x0 d x dx · f ′ x0 y0 · f ′ x0 f −1 ′ d tan−1 y dy sin−1 y . By the theorem of 1 f′ x y 1 cos x x x0 1 3 1 1 2 1 − sin x 1 − y2 1 f x0 ′ hx 4 sin 2x Lecture 4 - Differentiation Derivative of inverse function f −1 3 · h x0 . Derivative of inverse function ′ 3 ′ AMA140/150 Calculus f −1 1 2 − g h x0 x − x0 g h x − g h x0 h x − h x0 ′ gx x 1 g h x − g h x0 h x − h x0 lim fx 2 · h x0 . f x − f x0 lim x →x0 x − x0 lim 1 ′ Sketch of proof. Notice that h is differentiable at x0 and hence h is continuous at x0 . Therefore, h x → h x0 as x → x0 . Then f ′ x0 Example 4.6 Find the derivatives of the following functions. f −1 ′ AMA140/150 Calculus y0 1 . f ′ x0 Lecture 4 - Differentiation 1 d sin−1 y dy 1 1 − y2 2 d cos−1 y dy −1 3 1 − y2 AMA140/150 Calculus Lecture 4 - Differentiation d tan−1 y dy 1 1 y2 . Logarithmic function Properties of Logarithmic function Logarithmic function The logarithmic function f x ln x is defined as follows. 1 , t If x ≥ 1, then f x is the area enclosed by y t 1, and t x ; Properties of Logarithmic function the t -axis, the lines The logarithmic function ln x defined on 0, ∞ has the following properties. 1 the Notice that 1 ln 1 3 ln x < 0 if 0 < x < 1. ln a 2 a ln b ln b ; ln a − ln b ; ln an n ln a for all integer n. (Indeed, it holds for all real number n.) ln x > 0 if x > 1; 2 ln ab 3 If 0 < x < 1, the f x is the negative of the area enclosed by y t -axis, the lines t 1, and t x . 1 , t Derivative of Logarithmic function 0; 1 ln x is differentiable for all x ∈ 0, ∞ and d ln x dx 2 AMA140/150 Calculus ln x is a continuous function on 0, ∞ . Lecture 4 - Differentiation AMA140/150 Calculus Exponential function 1 . x Lecture 4 - Differentiation Properties of exponential function Exponential function Since the logarithmic function ln 0, ∞ → is one-to-one and onto inverse function exists. The inverse function is gx ex for all x ∈ , the . This function is called the exponential function. By the definition of inverse function, we have e ln x x for all x ∈ 0, ∞ , Properties of exponential function The exponential function satisfies the following properties. 1 e0 2 ea 3 The exponential function is differentiable everywhere and 1; b e a · e b for any real numbers a and b ; and ln e x x for all x ∈ d ex dx . Notice that e e 1 2.71828182845905 . . . . Sketch of proof of (3). Since the exponential function is the inverse of the e y . Applying logarithmic function, let y f x ln x . Then x f −1 y the result for derivative of inverse function on f , dy e dy AMA140/150 Calculus ex . Lecture 4 - Differentiation f −1 ′ y AMA140/150 Calculus 1 f′ x 1 1/x x ey . Lecture 4 - Differentiation Logarithmic function and exponential function Differentiation of parametric function Differentiation of parametric function Example 4.8 Find the derivatives of the following functions. 1 fx e sin x 2 gx ln x 2 3 ln ln cos x 2 hx Suppose x ft f′ x 2 g′ x 3 h′ x x2 dx dt provided that 2x − sin x . x 2 cos x 1 · 2x − sin x cos x Example 4.9 Find −2x sin x 2 . cos x 2 ln cos x 2 1 1 · · − sin x 2 · 2x ln cos x 2 cos x 2 dy dy and dx dx dx dt 3t 2 and 1 implies t Lecture 4 - Differentiation ft, y Example 4.10 Find gt, dy dx and dy dt 2t dy dx 2 dx provided that dt d dt dy dx dx dt 3x 2 x ,y dy if x 3 dx sin x cos x dy dx cos x y y cos x y − 3y 2 − 2y cos x , dx dt 3t 2 , dy dx − dy dt d dt 2 3t 2 ⇒ AMA140/150 Calculus 1,1 2 . 3 Lecture 4 - Differentiation y y3 y 2 cos x . 3y 2 dy dx 2y dy dx and t 2 for t > 0. d 2y dx 2 dy dx −2/3t 2 3t 2 dy dx 2 . 3t − Lecture 4 - Differentiation cos x y 2 − sin x dy dx −3x 2 − y 2 sin x − cos x −3x 2 − y 2 sin x − cos x y . cos x y − 3y 2 − 2y cos x In particular, 2t , Then 2 . 3t dy dx 2 t 3 and y 2t 3t 2 1. Then 0. dy Example 4.9 (cont.) Find if x dx 2 Solution. Notice that dy dx ⇒ Solution. Differentiate both sides by x , are differentiable functions of t . Then 2 t 2 for t > 0. 1,1 Implicit differentiation Differentiation of parametric function Suppose t 3 and y if x x ,y AMA140/150 Calculus Differentiation of parametric function , 0. dy dx AMA140/150 Calculus gt Solution. Now x , y x dy dt dx dt dy dx e sin x · cos x . y are differentiable functions of t . Then cos x Solution. 1 and x ,y 0,0 0−0−1 1−0−0 −1. 2 . 9t 4 AMA140/150 Calculus Lecture 4 - Differentiation y ...
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