Chapter 3 Notes, Applied Calculus, Tan

Transcription

1 Contents 3.1 Basic Rules of Differentiation The Prouct an Quotient Rules The Chain Rule Marginal Functions in Economics Higher Orer Derivatives Implicit Differentiation an relate rates

2 3.1 Basic Rules of Differentiation NOTATION The following are all ifferent ways of writing the erivative of y = f(x): f y (x), x, [f(x)] x, y We have four basic rules for taking erivatives. These four rules are provie without proof. Rule 1: Derivative of a constant x (c) = 0 Example Fin the erivatives of the following functions. a) y = 7 b) y = π Rule 2: The Power Rule if n is any real number, then x (xn ) = nx n 1 Example Fin the erivatives of the following functions. a) y = x 3 b) y = 3 x 2 Rule 3: Derivative of a Constant Multiple of a Function If c is a constant an f(x) is a ifferentiable function then x (cf(x)) = c x (f(x)) Example Fin the erivatives of the following functions. 2

3 a) y = 7x 3 b) y = 4 x 3 c) y = 1 4x 3 Rule 4: The Sum an Difference Rule If f(x) an g(x) are ifferentiable functions then [f(x) ± g(x)] = x x [f(x)] ± x [g(x)] Example Fin the erivatives of the following functions. a) y = 4x 5 + 3x 4 8x 2 + x + e 2 b) y = 7x (5x) 7 3

4 c) f(t) = 2t 2 t 3 Example Fin the tangent line to the curve at the given point. Put your answer in y = mx + b form. Point-slope form of the line with slope m passing through point (x 1, y 1 ): 1. f(x) = 2, (2, 2) y y 1 = m(x x 1 ) 2. f(x) = 4 x, (16, 2) 3. f(x) = 8 x 3, (2, 0) 4

5 4. f(x) = 2 9x 2, ( 1, 2 ) 9 Example Let f(x) = x 3 4x 2. Fin the point(s) on the graph of f where the tangent line is horizontal. Example The supply function for a certain make of satellite raio is given by p = f(x) = x 5/ where x is the quantity supplie an p is the unit price in ollars. a. Fin f (x). b. What is the rate of change of the unit price if the quantity supplie is 10,000 satelite raios? 5

6 3.2 The Prouct an Quotient Rules The Prouct Rule If f(x) an g(x) are ifferentiable functions then [f(x)g(x)] = f(x) [g(x)] + g(x) x x x [f(x)] It is often easier to think of this rule in terms of wors: The first times the erivative of the secon plus the secon times the erivative of the first. Example s(t) = (t 5 3t 2 + t)(t 4 3t 3 + 2t 2 t) Example f(x) = ( x 3 + 2x 2 + x ) ( x x ) The Quotient Rule If f(x) an g(x) are ifferentiable functions then x [ ] f(x) = g(x) g(x) [f(x)] f(x) x x [g(x)] [f(x)] 2 NOTE: There is a negative sign in this equation an yes, it oes matter which way you subtract. This one is also easier to remember in wors: The bottom times the erivative of the top minus the top times the erivative of the bottom all ivie by the bottom square. 6

7 Example f(t) = t2 t = top bottom Example g(x) = 3 x x + 1 = top bottom Rewrite this first. You nee exponents that are numbers. Example h(t) = 1 5t 2 7

8 Example f(x) = ( ) (2x 1x ) x 2 Example f(s) = s s 2 Example g(t) = (t + 1) (t2 + 1) t 2 8

9 3.3 The Chain Rule The chain rule is use for composition of functions: y = (f g)(x) = f(g(x)) There are two methos that can be use to solve for the erivative of a composition of functions. Both methos are calle the Chain Rule an are essentially the same using ifferent notation. Metho 1: If we rewrite the functions as y = f(u) an u = g(x) where both y an u are ifferentiable functions then we can take the erivatives separately: y u an If we multiply them u x together we will get an expression for y x : y x = y u u x Example Differentiate y = (x 5 8x) 2 using Metho 1. Inicate which function is y an which is u. Here we have a composition of functions. We can think of this as y = f(u) = u 2 an u = g(x) = x 5 8x If we take the composition of these two functions we get: 9

10 Q. Why use the chain rule? In the previous example we coul have easily multiplie out the function an taken a erivative. A. Because it makes y = (x 5 8x) 25 possible without multiplying it out. When we have small exponents there are few problems but when we have an exponent such as 25 it becomes more complicate. Example Differentiate y = (x 5 8x) 25 using Metho 1. Inicate which function is y an which is u. Metho 2: If y = (f g)(x) = f(g(x)) then y = f (g(x)) g (x) Both metho 1 an metho 2 are known as the chain rule. The secon metho may be easier. In the secon metho it is often easiest to think of these functions as the Outsie function an the Insie function. The outsie function is f(u) an the insie function is u = g(x). It is often easiest to remember the chain rule in wors: Take the erivative of the outsie function (leave the insie function alone) then multiply by the erivative of the insie function. 10

11 The chain rule can be written as: The Chain Rule If y = f(u) then y = f (u) u Using the chain rule we can write a more general form of the power rule. So far we can only take erivatives of a variable raise to an exponent: y = x n, but with the chain rule we can take the erivative of a function raise to an exponent. The General Power Rule Given the function then or y = [u(x)] n y x = n [u(x)]n 1 u x y = n u n 1 u Example y = 1 x

12 Example f(x) = 3x 2 x 5 Example y = x 3 (3x 2 9) 32 Sometimes you must combine two rules. Here we will use the prouct rule with the chain rule: Example y = x 4 x4 + 4 Here we will use the quotient rule with the chain rule: 12

13 ( t 2 Example y = t ) 3 13

14 3.4 Marginal Functions in Economics Marginal Cost Function If C(x) is the total cost function then the marginal cost function is C (x), the eravitive of the cost. The marginal cost represents the cost to make the next item. Marginal Revenue Function If p = f(x) is the unit price function then the revenue function R is given by R(x) = xf(x) an the marginal revenue function is given by R (x), the eravitive of the revenue. The marginal revenue represents the revenue realize from the sale of the (x+1) st item. Marginal Profit Function The profit function P is given by P (x) = R(x) C(x) an the marginal profit function is given by P (x), the eravitive of the profit. The marginal profit represents the profit realize from the sale of the (x+1) st item. Average Cost Function If C(x) is the total cost function then the average cost function enote by C(x) is C(x) = C(x) x 14

15 Example A ivision of Ditton Inustries manufactues the Futural moel microwave oven. The aily cost (in ollars) of proucing these microwave ovens is C(x) = x x x where x stans for the number of units prouce. 1. What is the actual cost incurre in manufacturing the 101st oven? the 201st oven? the 301st oven? 2. What is the marginal cost when x = 100, 200, an 300? 3. Fin the average cost function C(x). 4. Fin the marginal average cost function C (x) 15

16 Example The weekly eman for the Pulsar P television is p = x 0 x 12, 000 where p enotes the wholesale unit price (in ollars) an x enotes the quantity emane. The weekly total cost function associate with manufactuing the Pulsar 25 is given by C(x) = x x x + 80, 000 where C(x) enotes the total cost incurre in proucing x sets. 1. Fin the revenue function R(x) an the profit function P (x). 2. Fin the marginal cost function C (x), the marginal revenue function R (x), an the marginal profit function P (x). 3. Compute C (2000), R (2000), P (2000) an interpret your results. 16

17 3.5 Higher Orer Derivatives Example Fin the first an secon erivatives of the function. 1. f(x) = 0.2x 2 + 3x g(x) = 3x x f(x) = x 5 x 4 + 3x 3 + 3x 2 x 9 4. g(t) = t 2 (3t + 1) 4 17

18 5. h(w) = (w 2 + 2w + 4) 5/2 6. f(x) = 1 x Example Fin the thir erivative of g(t) = (3t 2 1) 5 18

19 3.6 Implicit Differentiation an relate rates Explicit Functions: Definition 3.1. An explicit function is a function in which one variable is efine only in terms of the other variable. (a) y = x Parabola (b) y = 9 x 2 Top half of the circle centere at the origin with raius 3. (c) y = 9 x 2 Bottom half of the circle centere at the origin with raius 3. Implicit Functions: Definition 3.2. An implicit function is a function in which one variable is not efine only in terms of another variable. (a) y x 2 = 7 Parabola (b) y 2 + x 2 = 9 Circle centere at the origin with raius 3. (c) y 4 + 7y 2 x y 2 x 4 9x 5 = 3 Some implicit functions can be written explicitly: both examples (a) represent the same parabola an (b) can be solve for either the top or the bottom of the circle. (c) can not be solve explicitly for y in terms of x. Implicit Differentiation: We have seen how to ifferentiate functions of the form y = f(x). We also want to be able to ifferentiate functions that either can t be written explicitly in terms of x or the resulting function is too complicate to eal with. To o this we use implicit ifferentiation. Key to implicit ifferentiation (chain rule): x [yn n 1 y ] = ny x an x [xn n 1 x ] = nx x = nxn 1 19

20 Steps to fin y x of implicit functions involving the variables x an y. 1. Treat y as a ifferentiable function of x. 2. Differentiate both sies of the equation with respect to x using all the rules we have previously use. i.e. Apply the operator to every term an use the prouct, quotient an chain rules. x 3. Solve for y x as if it were a variable. (a) Simplify each sie of the equal sign. (b) Bring all y x (c) Factor out y x. () Divie to isolate y x. terms to one sie of the equal sign an all non-y x terms to the other. Example Fin y x for x3 x 2 xy = 4 (a) by solving the equation explicitly in terms of y an (b) by implicit ifferentiation. 20

21 Example Fin y x for y2 + x 2 = 9 Example Fin y x for x2 y 2 + y 3 x = x 3 y 3 21