1 Definition of the derivative

Transcription

1 Math 20A - Calculus by Jon Rogawski Chapter 3 - Differentiation Prepare by Jason Gais Definition of the erivative Remark.. Recall our iscussion of tangent lines from way back. We now rephrase this in terms of limits. Let f(x) be a continuous function an P = (a, f(a)). If Q = (x, f(x)) is any other point on the graph of f, then the slope of the secant line is f x f(x) f(a) =. x a The closer x is to a, the closer this approximates the slope of the tangent line. Definition.2. The erivative of f(x) at x = a is the limit of the ifference quotients f f(x) f(a) (a), x a x a if this limit exists. When this limit exists, we say f is ifferentiable. Alternatively, taking x = a + h, we can rewrite the erivative as, f f(a + h) f(a) (a). Remark.3. The key point here is the following: the erivative at x = a is the slope of the tangent line at (a, f(a)). Definition.4. Assue f(x) is ifferentiable at x = a. The tangent line to the graph of y = f(x) at P = (a, f(a)) us the line through P of slope f (a). The equation of the tangent line is y f(a) = f (a)(x a). Example.5. Let f(x) = x 3. Write the equation of the tangent line to the graph y = f(x) at x =.

2 We compute the slope of the tangent line at x = by computing f (). f f(x) f() () x x x x 2 + x + = 3. x 3 x x (x )(x 2 + x + ) x x Thus, the equation of the tangent line is y = 3(x ), or equivalently, y = 3x 2. Example.6. Let f(x) = x 2. Compute f (2). To simplify matters, we first compute f(2 + h) f(2). f(2 + h) f(2) = Diviing by h gives (2 + h) 2 4 Now taking the limit, we get f(2 + h) f(2) h = 4 (2 + h)2 4(2 + h) 2. = = f (2) x 0 f(2 + h) f(2) h (4 + h) 4(2 + h) 2. 4h h2 (4 + h)h = 4(2 + h) 2 4(2 + h) 2 = 4 4(2) 2 = 4. 2

3 2 The erivative of a function Remark 2.. In the previous section, we compute the erivative at a point. This is inconvenient if we want to know the erivative of the same function at multiple points. In this section, we stuy the erivative function an look at ways to compute the erivative in several basic cases. Definition 2.2. We say a function f(x) is ifferentiable on (a, b) if the erivative exists for every point x (a, b). The function f f(x + h) f(x) (x), is the erivative of f. Alternatively, we write (in Leibniz notation), f (x) = x f(x). Example 2.3. Compute the erivative of f(x) = x + x. f f(x + h) f(x) (x + h) + x + h (x + x) (x) h 0 h ( ) h + x + h x x + h x + h 0 h x + h x = + lim ( x + h + x) = + lim = + h 0 x + h + x 2 x. Remark 2.4 (Power Rule). It is worth reaing the proof of the following rule on page 3, x xn = nx n. Example 2.5. Compute the erivative of f(x) = x + x. f (x) = x x + x x/2 = x x /2 = + 2 x. 3

4 Remark 2.6 (Linearity Rules). Let f an g be ifferentiable functions an c a constant. Then x (f + g) = f + g an x cf = cf. We prove the secon. The proof of the first is on page 32. cf(x + h) cf(x) cf(x) x f(x + h) f(x) = c lim f(x + h) f(x) c = cf (x). Example 2.7. Compute the erivative of f(x) = e x. f f(x + h) f(x) (x) = e x e h lim = e x. This last step follows by the efinition of e x. Theorem 2.8. Differentiability implies continuity. Proof. Suppose f is ifferentiable at x = c. x = c. That is, lim f(x) = f(c). x c For x c, we have f(x) f(c) = (x c) On either sie, we compute the limit at x c. ( (x c) On the other han, lim(f(x) f(c)) x c x c h 0 e x+h e x h We claim f is continuous at f(x) f(c). x c f(x) f(c) x c ) f(x) f(c) (x c) lim x c x c x c = 0 f (c) = 0. lim(f(x) f(c)) f(x) lim f(c) f(x) f(c). x c x c x c x c The result now follows. 4

5 Remark 2.9. The converse to this statement is not true. Let f(x) = x, then f f(x) f(0) (0) x 0 x 0 x x 0 x. We have seen before that this limit oes not exist. Remark 2.0. Let f(x) = x n with 0 < n <, then f is not ifferentiable at x = 0. In particular, the tangent line is vertical. 5

6 3 Prouct an quotient rules Remark 3.. We stuy two aitional rules that allow us to compute certain erivatives quickly. Theorem 3.2 (Prouct rule). Suppose f an g are ifferentiable, then fg is ifferentiable an Proof. (fg) (x) = f (x)g(x) + f(x)g (x). (fg) f(x + h)g(x + h) f(x)g(x) (x) f(x + h)g(x + h) f(x + h)g(x) + f(x + h)g(x) f(x)g(x) f(x + h)(g(x + h) g(x)) + (f(x + h) f(x)) g(x) g(x + h) g(x) f(x + h) f(x) f(x + h) + lim g(x) = f(x)g (x) + f (x)g(x). Example 3.3. Let h(x) = e x (5x 2 + x ). Compute h (x). Theorem 3.4 (Quotient rule). Suppose f an g are ifferentiable, then f/g is ifferentiable for all x with g(x) 0 an ( ) f (x) = f (x)g(x) f(x)g (x) g (g(x)) 2. Remark 3.5. This might be best remembere in the following way: ( ) hi = lo (lo)(hi) (hi)(lo). lolo Example 3.6. Let h(x) = x2 +3x 3 2x 5x 2. Compute h (x). 6

7 4 Rates of Change Remark 4.. Recall that if y = f(x) is a function over an interval [x 0, x ], then we enote the change in f as f = f(x ) f(x 0 ) an change in x as x = x x 0. Then average rate of change is given as f x an instantaneous f rate of change is given as lim x 0 x. Example 4.2. Let A = s 3 be the area of a cube of sie length s cm. Compute the rate of change of A with respect to s when s = 4. We compute the erivative A s = s (s3 ) = 3s 2 A s = 3(4) 2 = 48cm/sec. s=4 Note that, when s = 5 then the rate is 75 cm/sec. Hence, the larger the sie length, the faster the rate of growth. This is because rate of growth epens on the length of the sie. Remark 4.3. Recall that velocity is the change in position with respect to time. That is, velocity is the instantaneous rate of change of the position function. That is, v(t) = s t where s(t) is the position function. The sign of v(t) (±) inicates irection (up/own, forwar/backwar). Spee is the absolute value of the velocity function. The formulas (iscovere by Galileo) for height an velocity of an object thrown irectly into the air (near the surface of the earth) are given as s(t) = s 0 + v 0 t 2 gt2, v(t) = s t = v 0 gt, where s 0 = s(0) an v 0 = v(0). Notice that v(t) = s (t). The constant g is acceleration ue to gravity, so g 9.8m/sec 2 or g 32ft/sec 2. Example 4.4. A ball is fire irectly into the air with an initial velocity of 64ft/sec from an initial height of 20ft. Fin the maximum height. The maximum height occurs when v(t) = 0. We have Thus, the position is 0 = v(t) = 64 32t t = 2. s(2) = (2) 6(2 2 ) = 84ft. 7

8 Remark 4.5 (Marginal Cost). If C(x) is the cost function for proucing x units, then marginal cost is efine as the cost of proucing one aitional unit. That is, C(x 0 + ) C(x 0 ). However, we can approximate marginal cost using the erivative C (x 0 ). Example 4.6. Let C(x) = x 0.5(x/000) 3 be the cost of proucing x bagels (in ollars). Compute the cost of proucing 2000 bagels. Compare the estimate an exact value of the marginal cost of proucing 200 bagels. We have C(2000) = 796. Now C (x) =.25 (.5/000 3 )x 2. Then C (200) The actual marginal cost is C(200) C(2000) = Higher erivatives 8

9 6 Trig functions Remark 6.. We ll start toay by working out the erivative of cos x with the limit efinition of the erivative. cos(x + h) cos x cos x. x Using the angle sum ientity for cos(x + h) we have Thus, we have cos(x + h) cos x = cos x cos h sin x sin h cosx = cos x(cos h ) sin x sin h. cos(x + h) cos x cos x x h 0 ( h ) cos x(cos h ) sin x sin h h cos h sin h = cos x lim sin x lim = (cos x) 0 (sin x) = sin x. Theorem 6.2. The functions y = sin x an y = cos x are ifferentiable an sin x = cos x an cos x = sin x. x x Remark 6.3. The rules for the other four trig functions can now be worke out with the quotient rule. Prove rule for sec x. Theorem 6.4. x tan x = sec2 x, x cot x = csc2 x, sec x = sec x tan x x csc x = csc x cot x. x 9

10 7 The chain rule Theorem 7.. If f an g are ifferentiable, then the composite function (f g)(x) = f(g(x)) is ifferentiable an (f(g(x)) = f (g(x)) g (x). Example 7.2. Compute the erivative of sin(cos(x)). Example 7.3. Compute the erivative of x Theorem 7.4 (General power an exponential rules). If g(x) is ifferentiable, then x g(x)n = ng(x) n g (x) x e( g(x)) = g (x)e g(x). Example 7.5. Compute the erivative of tan(e cos x ). 0

11 8 Derivatives of inverse functions Example 8.. Let f(x) = x on [0, ]. Compute the erivative of f. First, we fin the inverse function g(x) = f (x). Algebra shows g(x) = x 4. Then, g (x) = 2 (x 4) /2. Theorem 8.2 (Derivative of the Inverse). Assume that f(x) is ifferentiable an - with inverse g(x) = f (x). If b belongs to the omain of g(x) an f (g(b)) 0, then g (b) exists an g (b) = f (g(b)). Remark 8.3. The proof that g(x) is ifferentiable is beyon the scope of this class. In particular, it uses the formal efinition of a limit. However, if we assume that g(x) is ifferentiable, then we can use the chain rule to erive the formula above. Since g(x) is the inverse of f(x), then f(g(x)) = x. Differentiating both sies gives x f(g(x)) = x x f (g(x)) g (x) = g (x) = /f (g(x)). Graphically, we recall that the graph of g is the reflection of the graph of f over the line y = x. Thus, the slopes of the tangent lines at a point shoul be reciprocals of one another. Example 8.4. This result matches our previous example. Since f (x) = 2x. Then f (g(x)) = 2(x 4) /2 an so g (x) = /f (g(x)) = 2 x 4. Example 8.5. Compute the erivative of sin x an cos x. Let f(x) = sin x an g(x) = sin x. Then f (x) = cos x an an so g (x) = /f (g(x)) = / cos(sin (x)). Consier the right triangle with leg x an hypotenuse. The other leg then has length x 2 an so cos sin (x) = x 2 so x sin x = x 2.

12 The book uses the trick that cos x + sin x = π 2 to compute x cos x. However, it is easy enough to work out in a manner similar to the above, giving x cos x =. x 2 Theorem 8.6. The erivatives for the six inverse trig functions are given below, x sin x = x 2 x cos x = x 2 x tan x = + x 2 x cot x = + x 2 x sec x = x x 2 x csc x = x x 2 2

13 9 Derivative of general exponential an logarithmic functions Theorem 9.. Let f(x) = b x for b > 0. Then Proof. By the chain rule, x bx = (ln b)b x. x bx = x eln bx = x ex ln b = (ln b)e x ln b = (ln b)b x. Example 9.2. Compute the erivative of f(x) = 2 cos x. Theorem 9.3. For x > 0, x ln x = x. Proof. Let f(x) = e x so f (x) = ln x. Then x ln x = x f (x) = f (ln x) = e = ln x x. Remark 9.4. The chain rule now implies x ln(f(x)) = f (x) f(x). 3

14 0 Implicit ifferentiation Remark 0.. Suppose we want the equation for the tangent line on a circle at a given point. The equation of a circle is x 2 + y 2 =. So to get the erivative we have to solve for y an then take the erivative (technically, two erivatives). (Do this!) An easier way is to use implicit ifferentiation. function of x an apply the chain rule. That is, we treat y as a Example 0.2. Fin y/x with x 2 + y 2 =. Use it to fin the equation of the tangent line at ( 2, 3 2 ). We treat y/x as an unknown. ( x 2 + y 2) = x x () x (x2 ) + x (y2 ) = 0 2x + 2y y = 0 by the chain rule x y x = x y. Remark 0.3. It may not always be possible to solve for y easily an so implicit ifferentiation becomes an important tool. Example 0.4. Fin y/x with xe y = 2xy + y 3. Remark 0.5. If we are only intereste in the erivative at a point, then it is not necessary to solve for y/x. Example 0.6. Fin the equation of the tangent line at the point (, 0) for the function x 2 + sin y = xy

15 Relate Rates Remark.. Suppose we have a function which changes accoring to two variables. We can use implicit ifferentiation to compare rates for both variables, treating them both as inepenent variables. Example.2. Each sie of a square is increasing at a rate of 6 cm/sec. At what rate is the area of the square increasing when the area of the square is 6 cm 2? We have A = s 2, so A t = 2s s A t. When A = 6, s = 4, so t = 2(4)(6) = 48 cm 2 /sec when s t = 6 cm/sec. Example.3. A spotlight on the groun shines on a wall 2 m away. If a man 2 m tall walks from the spotlight towar the builing at a spee of.6 m/sec, how fast is the length of his shaow on the builing ecreasing when he is 4 m from the builing? See note page. Example.4. Water is leaking out of an inverte conical tank at a rate of 0,000 cm 3 /min at the same time that water is being pumpe into the tank at a constant rate. The tank has height 6m an the iameter at the top is 4m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2m, fin the rate at which water is being pumpe into the tank. If C is the rate the water is pumpe in, then V t = C 0, 000, where V = 3 πr2 h is the volume at time t. By similar triangles, Thus, When h = 2 m, h t r 2 = h 6 r = 3 h. V = 3 π ( 3 h ) 2 h = π 27 h3 V t = π 9 h2h t. = 20 cm/min, so C 0, 000 = π 9 (200)2 (20) C = 0, , 000 π 289, 253cm 3 /min. 9 5