MIT单变量微积分unit1-8

MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 1 18.01 Fall 2006 Unit 1: Derivatives A. What is a derivative? • Geometric interpretation • Physical interpretation • Important for any measurement (economics, political science, finance, physics, etc.) B. How to differentiate any function you know. d � � For example: e x arctan x . We will discuss what a derivative is today. Figuring out how to • dx differentiate any function is the subject of the first two weeks of this course. Lecture 1: Derivatives, Slope, Velocity, and Rate of Change Geometric Viewpoint on Derivatives Tangent line Secant line f(x) P Q x0 x0+∆x y Figure 1: A function with secant and tangent lines The derivative is the slope of the line tangent to the graph of f(x). But what is a tangent line, exactly? 1 Lecture 1 18.01 Fall 2006 • It is NOT just a line that meets the graph at one point. • It is the limit of the secant line (a line drawn between two points on the graph) as the distance between the two points goes to zero. Geometric definition of the derivative: Limit of slopes of secant lines PQ as Q P (P fixed). The slope of PQ:→ P Q (x0+∆x, f(x0+∆x)) (x0, f(x0)) ∆x ∆f Secant Line Figure 2: Geometric definition of the derivative lim Δf = lim f(x0 +Δx) − f(x0) = f �(x0) Δx 0 Δx Δx 0 Δx � �� �→ → � �� � “difference quotient” “derivative of f at x0 ” 1 Example 1. f(x) = x One thing to keep in mind when working with derivatives: it may be tempting to plug in Δx = 0 Δf 0 right away. If you do this, however, you will always end up with = . You will always need to Δx 0 do some cancellation to get at the answer. Δf 1 1 1 � x0 − (x0 +Δx) � 1 � � = x0 +Δx − x0 = = −Δx = −1 Δx Δx Δx (x0 +Δx)x0 Δx (x0 +Δx)x0 (x0 +Δx)x0 Taking the limit as Δx 0,→ lim −1 = −1 Δx→0 (x0 +Δx)x0 x20 2 Lecture 1 18.01 Fall 2006 y x x0 Figure 3: Graph of x 1 Hence, f �(x0) = − 2 1 x0 Notice that f �(x0) is negative — as is the slope of the tangent line on the graph above. Finding the tangent line. Write the equation for the tangent line at the point (x0, y0) using the equation for a line, which you all learned in high school algebra: y − y0 = f �(x0)(x − x0) Plug in y0 = f(x0) = 1 and f �(x0) = − 2 1 to get: x0 x0 y − x 1 0 = − x20 1 (x − x0) 3 Lecture 1 18.01 Fall 2006 y x x0 Figure 4: Graph of x 1 Just for fun, let’s compute the area of the triangle that the tangent line forms with the x- and y-axes (see the shaded region in Fig. 4). First calculate the x-intercept of this tangent line. The x-intercept is where y = 0. Plug y = 0 into the equation for this tangent line to get: 0 − 1 = −2 1 (x − x0) x0 x0 −1 −1 1 = x +2x0 x0 x0 1 2 x = 2x0 x0 2 x = x 20( ) = 2x0 x0 So, the x-intercept of this tangent line is at x = 2x0. 1 1 Next we claim that the y-intercept is at y = 2y0. Since y = and x = are identical equations, x y the graph is symmetric when x and y are exchanged. By symmetry, then, the y-intercept is at y = 2y0. If you don’t trust reasoning with symmetry, you may follow the same chain of algebraic reasoning that we used in finding the x-intercept. (Remember, the y-intercept is where x = 0.) Finally, 1 1 Area = (2y0)(2x0) = 2x0y0 = 2x0( ) = 2 (see Fig. 5) 2 x0 Curiously, the area of the triangle is always 2, no matter where on the graph we draw the tangent line. 4 Lecture 1 18.01 Fall 2006 y x x0 2x0 y0 2y0 x-1 Figure 5: Graph of x 1 Notations Calculus, rather like English or any other language, was developed by several people. As a result, just as there are many ways to express the same thing, there are many notations for the derivative. Since y = f(x), it’s natural to write Δy = Δf = f(x) − f(x0) = f(x0 +Δx) − f(x0) We say “Delta y” or “Delta f” or the “change in y”. If we divide both sides by Δx = x − x0, we get two expressions for the difference quotient: Δy = Δf Δx Δx Taking the limit as Δx → 0, we get Δy Δx → dy dx (Leibniz’ notation) Δf Δx → f �(x0) (Newton’s notation) When you use Leibniz’ notation, you have to remember where you’re evaluating the derivative — in the example above, at x = x0. Other, equally valid notations for the derivative of a function f include df , f �, and Df dx 5 Lecture 1 18.01 Fall 2006 Example 2. f(x) = x n where n = 1, 2, 3... d What is x n? dx To find it, plug y = f(x) into the definition of the difference quotient. n nΔy = (x0 +Δx)n − x0 = (x +Δx) n − x Δx Δx Δx (From here on, we replace x0 with x, so as to have less writing to do.) Since (x +Δx)n = (x +Δx)(x +Δx)...(x +Δx) n times We can rewrite this as � � x n + n(Δx)x n−1 + O (Δx)2 O(Δx)2 is shorthand for “all of the terms with (Δx)2, (Δx)3, and so on up to (Δx)n.” (This is part of what is known as the binomial theorem; see your textbook for details.) n nΔy = (x +Δx)n − x = xn + n(Δx)(xn−1) + O(Δx)2 − x = nx n−1 + O(Δx) Δx Δx Δx Take the limit: Δy lim = nx n−1 Δx 0 Δx→ Therefore, d n x = nx n−1 dx This result extends to polynomials. For example, d 9(x 2 + 3x 10) = 2x + 30x dx Physical Interpretation of Derivatives You can think of the derivative as representing a rate of change (speed is one example of this). On Halloween, MIT students have a tradition of dropping pumpkins from the roof of this building, which is about 400 feet high. The equation of motion for objects near the earth’s surface (which we will just accept for now) implies that the height above the ground y of the pumpkin is: y = 400 − 16t2 Δy distance travelled The average speed of the pumpkin (difference quotient) = = Δt time elapsed When the pumpkin hits the ground, y = 0, 400 − 16t2 = 0 6 Lecture 1 18.01 Fall 2006 Solve to find t = 5. Thus it takes 5 seconds for the pumpkin to reach the ground. 400 ft Average speed = = 80 ft/s 5 sec A spectator is probably more interested in how fast the pumpkin is going when it slams into the ground. To find the instantaneous velocity at t = 5, let’s evaluate y�: y� = −32t = (−32)(5) = −160 ft/s (about 110 mph) y� is negative because the pumpkin’s y-coordinate is decreasing: it is moving downward. 7 Lecture 2 18.01 Fall 2006 Lecture 2: Limits, Continuity, and Trigonometric Limits More about the “rate of change” interpretation of the derivative y = f(x) y x ∆x ∆y Figure 1: Graph of a generic function, with Δx and Δy marked on the graph 3. T temperature gradient Δy Δx → dy dx as Δx → 0 Average rate of change → Instantaneous rate of change Examples 1. q = charge dq dt = electrical current 2. s = distance ds dt = speed dT = temperature = dx 1 Lecture 2 18.01 Fall 2006 4. Sensitivity of measurements: An example is carried out on Problem Set 1. In GPS, radio signals give us h up to a certain measurement error (See Fig. 2 and Fig. 3). The question is ΔL how accurately can we measure L. To decide, we find . In other words, these variables are Δh related to each other. We want to find how a change in one variable affects the other variable. L hs satellite you Figure 2: The Global Positioning System Problem (GPS) hs L Figure 3: On problem set 1, you will look at this simplified “flat earth” model 2 � Lecture 2 18.01 Fall 2006 Limits and Continuity Easy Limits x2 + x 32 + 3 12 lim = = = 3 x→3 x + 1 3 + 1 4 With an easy limit, you can get a meaningful answer just by plugging in the limiting value. Remember, lim Δf = lim f(x0 +Δx) − f(x0) x→x0 Δx x→x0 Δx is never an easy limit, because the denominator Δx = 0 is not allowed. (The limit x x0 is computed under the implicit assumption that x =� x0.) → Continuity We say f(x) is continuous at x0 when lim f(x) = f(x0) x x0→ Pictures x y Figure 4: Graph of the discontinuous function listed below x + 1 x > 0 f(x) = −x x ≥ 0 3 Lecture 2 18.01 Fall 2006 This discontinuous function is seen in Fig. 4. For x > 0, lim f(x) = 1 x 0→ but f(0) = 0. (One can also say, f is continuous from the left at 0, not the right.) 1. Removable Discontinuity Figure 5: A removable discontinuity: function is continuous everywhere, except for one point Definition of removable discontinuity Right-hand limit: lim f(x) means lim f(x) for x > x0. + 0 x x0→x x→ Left-hand limit: lim f(x) means lim f(x) for x < x0. 0x − f(x) = lim f(x) but this is not f(x0), or if f(x0) is undefined, we say the disconti­ x x0x →→ If lim + 0 0x −→ nuity is removable. x x x→ For example, sin( x x) is defined for x = 0. We will see later how to evaluate the limit as � x → 0. 4 Lecture 2 18.01 Fall 2006 2. Jump Discontinuity x0 Figure 6: An example of a jump discontinuity lim for (x < x0) exists, and lim for (x > x0) also exists, but they are NOT equal. + 0 −x0x x x→ → 3. Infinite Discontinuity y x Figure 7: An example of an infinite discontinuity: 1 x 1 1 Right-hand limit: lim = ∞; Left-hand limit: lim x→0+ x x→0− x = −∞ 5 Lecture 2 18.01 Fall 2006 4. Other (ugly) discontinuities Figure 8: An example of an ugly discontinuity: a function that oscillates a lot as it approaches the origin This function doesn’t even go to ±∞ — it doesn’t make sense to say it goes to anything. For something like this, we say the limit does not exist. 6 Lecture 2 18.01 Fall 2006 Picturing the derivative x y x y’ Figure 9: Top: graph of f (x) = 1 and Bottom: graph of f �(x) = − 1 2 x x Notice that the graph of f(x) does NOT look like the graph of f �(x)! (You might also notice that f(x) is an odd function, while f �(x) is an even function. The derivative of an odd function is always even, and vice versa.) 7 Lecture 2 18.01 Fall 2006 Pumpkin Drop, Part II This time, someone throws a pumpkin over the tallest building on campus. Figure 10: y = 400 − 16t2 , −5 ≤ t ≤ 5 Figure 11: Top: graph of y(t) = 400 − 16t2 . Bottom: the derivative, y�(t) 8 Lecture 2 18.01 Fall 2006 Two Trig Limits Note: In the expressions below, θ is in radians— NOT degrees! lim sin θ = 1; lim 1 − cos θ = 0 θ 0 θ θ 0 θ→ → Here is a geometric proof for the first limit: 1 θ arc length = θ sinθ Figure 12: A circle of radius 1 with an arc of angle θ sin θ arc length = θ 1 θ Figure 13: The sector in Fig. 12 as θ becomes very small Imagine what happens to the picture as θ gets very small (see Fig. 13). As θ 0, we see that sin θ → 1. θ → 9 Lecture 2 18.01 Fall 2006 What about the second limit involving cosine? 1 cos θ 1 - cos θ arc length = θ θ Figure 14: Same picture as Fig. 12 except that the horizontal distance between the edge of the triangle and the perimeter of the circle is marked From Fig. 15 we can see that as θ → 0, the length 1 − cos θ of the short segment gets much smaller than the vertical distance θ along the arc. Hence, 1 − cos θ 0. θ → 1 cos θ 1 - cos θ arc length = θ θ Figure 15: The sector in Fig. 14 as θ becomes very small 10 � � Lecture 2 18.01 Fall 2006 We end this lecture with a theorem that will help us to compute more derivatives next time. Theorem: Differentiable Implies Continuous. If f is differentiable at x0, then f is continuous at x0. f(x) − f(x0)Proof: x lim x0 (f(x) − f(x0)) = x lim x0 x − x0 (x − x0) = f �(x0) · 0 = 0. → → Remember: you can never divide by zero! The first step was to multiply by x − x0 . It looks as 0 x − x0 if this is illegal because when x = x0, we are multiplying by . But when computing the limit as 0 x → x0 we always assume x �= x0. In other words x − x0 �= 0. So the proof is valid. 11 � � Lecture 3 18.01 Fall 2006 Lecture 3 Derivatives of Products, Quotients, Sine, and Cosine Derivative Formulas There are two kinds of derivative formulas: d d 1 1. Specific Examples: x n or dx dx x 2. General Examples: (u + v)� = u� + v� and (cu) = cu� (where c is a constant) A notational convention we will use today is: (u + v)(x) = u(x) + v(x); uv(x) = u(x)v(x) Proof of (u + v) = u� + v�. (General) Start by using the definition of the derivative. (u + v)�(x) = lim (u + v)(x +Δx) − (u + v)(x) Δx 0 Δx→ = lim u(x +Δx) + v(x +Δx) − u(x) − v(x) Δx 0 Δx→ � � = lim u(x +Δx) − u(x) + v(x +Δx) − v(x) Δx 0 Δx Δx→ (u + v)�(x) = u�(x) + v�(x) Follow the same procedure to prove that (cu)� = cu�. Derivatives of sin x and cos x. (Specific) Last time, we computed sin x lim = 1 x→0 x d (sin x) x=0 = lim sin(0 + Δx) − sin(0) = lim sin(Δx) = 1 dx | Δx→0 Δx Δx→0 Δx d (cos x) x=0 = lim cos(0 + Δx) − cos(0) = lim cos(Δx) − 1 = 0 dx | Δx→0 Δx Δx→0 Δx d d So, we know the value of sin x and of cos x at x = 0. Let us find these for arbitrary x. dx dx d sin x = lim sin(x +Δx) − sin(x) dx Δx 0 Δx→ 1 � � � � Lecture 3 18.01 Fall 2006 Recall: sin(a + b) = sin(a) cos(b) + sin(b) cos(a) So, d sin x = lim sin x cosΔx + cos x sinΔx − sin(x) dx Δx 0 Δx→ � � = lim sin x(cosΔx − 1) + cos x sinΔx Δx 0 Δx Δx→ � � � � = lim sin x cosΔx − 1 + lim cos x sinΔx Δx 0 Δx Δx 0 Δx→ → Since cos Δx − 1 0 and that sin Δx 1, the equation above simplifies to Δx → Δx → d sin x = cos x dx A similar calculation gives d cos x = − sin x dx Product formula (General) (uv)� = u�v + uv� Proof: (uv)� = lim (uv)(x +Δx) − (uv)(x) = lim u(x +Δx)v(x +Δx) − u(x)v(x) Δx 0 Δx Δx 0 Δx→ → Now obviously, u(x +Δx)v(x) − u(x +Δx)v(x) = 0 so adding that to the numerator won’t change anything. (uv)� = lim u(x +Δx)v(x) − u(x)v(x) + u(x +Δx)v(x +Δx) − u(x +Δx)v(x) Δx 0 Δx→ We can re-arrange that expression to get (uv)� = lim u(x +Δx) − u(x) v(x) + u(x +Δx) v(x +Δx) − v(x) Δx 0 Δx Δx→ Remember, the limit of a sum is the sum of the limits. � � � � �� lim u(x +Δx) − u(x) v(x) + lim u(x +Δx) v(x +Δx) − v(x) Δx 0 Δx Δx 0 Δx→ → (uv)� = u�(x)v(x) + u(x)v�(x) Note: we also used the fact that lim u(x +Δx) = u(x) (true because u is continuous) Δx 0→ This proof of the product rule assumes that u and v have derivatives, which implies both functions are continuous. 2 Lecture 3 18.01 Fall 2006 u ∆u ∆v v Figure 1: A graphical “proof” of the product rule An intuitive justification: We want to find the difference in area between the large rectangle and the smaller, inner rectangle. The inner (orange) rectangle has area uv. Define Δu, the change in u, by Δu = u(x +Δx) − u(x) We also abbreviate u = u(x), so that u(x + Δx) = u + Δu, and, similarly, v(x + Δx) = v + Δv. Therefore the area of the largest rectangle is (u +Δu)(v +Δv). If you let v increase and keep u constant, you add the area shaded in red. If you let u increase and keep v constant, you add the area shaded in yellow. The sum of areas of the red and yellow rectangles is: [u(v +Δv) − uv] + [v(u +Δu) − uv] = uΔv + vΔu If Δu and Δv are small, then (Δu)(Δv) ≈ 0, that is, the area of the white rectangle is very small. Therefore the difference in area between the largest rectangle and the orange rectangle is approximately the same as the sum of areas of the red and yellow rectangles. Thus we have: [(u +Δu)(v +Δv) − uv] ≈ uΔv + vΔu (Divide by Δx and let Δx 0 to finish the argument.) → 3 Lecture 3 18.01 Fall 2006 Quotient formula (General) To calculate the derivative of u/v, we use the notations Δu and Δv above. Thus, u(x +Δx) u(x) = u +Δu u v(x +Δx) − v(x) v +Δv − v = (u +Δu)v − u(v +Δv) (common denominator) (v +Δv)v = (Δu)v − u(Δv) (v + Δv)v (cancel uv − uv) Hence, 1 Δx � u + Δu v + Δv − u v � = ( Δu Δx )v − u( Δv Δx ) (v + Δv)v −→ v( du dx ) − u( dv dx ) v2 as Δx → 0 Therefore, ( u v )� = u�v − uv� v2 . 4 Lecture 4 Sept. 14, 2006 18.01 Fall 2006 Lecture 4 Chain Rule, and Higher Derivatives Chain Rule We’ve got general procedures for differentiating expressions with addition, subtraction, and multi­ plication. What about composition? Example 1. y = f(x) = sin x, x = g(t) = t2 . So, y = f(g(t)) = sin(t2). To find dy , write dt t = t0 +Δtt0 = t0 x = x0 +Δxx0 = g(t0) y = y0 +Δyy0 = f(x0) Δy = Δy Δx Δt Δx · Δt As Δt 0, Δx 0 too, because of continuity. So we get: → → dy dy dx = The Chain Rule! dt dx dt ← In the example, dx dt = 2t and dy dx = cos x. So, d dt � sin(t2) � = ( dy dx )( dx dt ) = = (cos x)(2t) (2t) � cos(t2) � Another notation for the chain rule � � d dt f(g(t)) = f �(g(t))g�(t) or d dx f(g(x)) = f �(g(x))g�(x) Example 1. (continued) Composition of functions f(x) = sin x and g(x) = x2 (f g)(x) = f(g(x)) = sin(x 2)◦ (g f)(x) = g(f(x)) = sin2(x)◦ Note: f ◦ g �= g ◦ f. Not Commutative! 1 � � � � � � Lecture 4 Sept. 14, 2006 18.01 Fall 2006 x g g(x) f(g(x))f Figure 1: Composition of functions: f g(x) = f(g(x))◦ d 1 Example 2. cos =? dx x 1 Let u = x dy = dy du dx du dx dy du 1 du = − sin(u); dx = − x2 � � 1 � � sin dy sin(u) x = = (− sin u) −1 = dx x2 x2 x2 d � � Example 3. x−n =? dx � �n1 1 There are two ways to proceed. x−n = , or x−n = x xn 1. d � x−n � = d � 1 �n = n � 1 �n−1 � −1 � = −nx−(n−1)x−2 = −nx−n−1 dx dx x x x2 2. d � x−n � = d 1 = nx n−1 −1 = −nx−n−1 (Think of xn as u) dx dx xn x2n 2 � � Lecture 4 Sept. 14, 2006 18.01 Fall 2006 Higher Derivatives Higher derivatives are derivatives of derivatives. For instance, if g = f �, then h = g� is the second derivative of f . We write h = (f �)� = f ��. Notations f �(x) f ��(x) f ���(x) f (n)(x) Df D2f D3f Dnf df dx d2f dx2 d3f dx3 dnf dxn Higher derivatives are pretty straightforward —- just keep taking the derivative! nExample. Dnx =? Start small and look for a pattern. Dx = 1 D2 x 2 = D(2x) = 2 (= 1 2) · D3 x 3 = D2(3x 2) = D(6x) = 6 (= 1 2 3)· · D4 x 4 = D3(4x 3) = D2(12x 2) = D(24x) = 24 (= 1 2 3 4)· · · Dn x n = n! we guess, based on the pattern we’re seeing here. ← The notation n! is called “n factorial” and defined by n! = n(n − 1) 2 1· · · · Proof by Induction: We’ve already checked the base case (n = 1). nInduction step: Suppose we know Dnx = n! (nth case). Show it holds for the (n + 1)st case. Dn+1 x n+1 = Dn Dxn+1 = Dn ((n + 1)x n) = (n + 1)Dn x n = (n + 1)(n!) Dn+1 x n+1 = (n + 1)! Proved! 3 � � � � � � Lecture 5 18.01 Fall 2006 Lecture 5 Implicit Differentiation and Inverses Implicit Differentiation d Example 1. (x a) = ax a−1 . dx We proved this by an explicit computation for a = 0, 1, 2, .... From this, we also got the formula for a = −1, −2, .... Let us try to extend this formula to cover rational numbers, as well: m m a = ; y = x n where m and n are integers. n We want to compute dy . We can say yn = xm so nyn−1 dy = mx m−1 . Solve for dy : dx dx dx dy = m xm−1 dx n yn−1 ( m We know that y = x n ) is a function of x. dy = m xm−1 dx n yn−1 m xm−1 = n (xm/n)n−1 m xm−1 = n xm(n−1)/n = x(m−1)− m(n n −1)m n m n(m−1)−m(n−1) = x n n m nm−n−nm+m = x n n m m n = x n − n n dy m m So, = x n − 1 dx n This is the same answer as we were hoping to get! Example 2. Equation of a circle with a radius of 1: x2 +y2 = 1 which we can write as y2 = 1−x2 . So y = ±√1 − x2. Let us look at the positive case: � 1 y = + 1 − x2 = (1 − x 2) 2 dy = 1 (1 − x 2) −21 (−2x) = −x = −x dx 2 √ 1 − x2 y 1 Lecture 5 18.01 Fall 2006 Now, let’s do the same thing, using implicit differentiation. x 2 + y 2 = 1 d � 2� d x 2 + y = (1) = 0 dx dx d d (x 2) + (y 2) = 0 dx dx Applying chain