Chapter 4 Notes, Calculus I with Precalculus 3e Larson/Edwards

Transcription

1 4.1 The Derivative Recall: For the slope of a line we need two points (x 1,y 1 ) and (x 2,y 2 ). Then the slope is given by the formula: m = y x = y 2 y 1 x 2 x 1 On a curve we can find the slope of a Secant Line because a secant passes through the curve at least two times. The points of intersection are (x 1,f(x 1 )) and (x 2,f(x 2 )). So the slope can be written: m = y x = f(x 2) f(x 1 ) x 2 x 1. Or if we let x 2 = x 1 +h then the equation becomes m = y x = f(x 1 +h) f(x 1 ). h What about the slope of the tangent line at x 1? 1

2 Now we dont have two points to find the slope we only know (x 1,f(x 1 )). So what can we do? We can find another point on the curve that is CLOSE to (x 1,f(x 1 )) and use that to find the slope. How CLOSE? lim h 0 CLOSE. To find the slope of the tangent line at x 1 we take the limit of the slope of the secant line: y lim h 0 x = lim h 0 f(x 1 +h) f(x 1 ). h If we do this in general for a generic x then we get the DERIVATIVE of the function. The Derivative is a function (denoted f (x) ) which will tell you the SLOPE OF THE TANGENT LINE at any point on the function. It is given by the following formula: f f(x+h) f(x) (x) = lim. h 0 h Example Find the slope of the tangent line to f(x) = 5 x 2 at the point (2,1) In this case (x 1,f(x 1 )) = (2,1) 2

3 Example Find the derivative of f(x) = x 3 +x 2 and the tangent line at the point (1,2). Example Let f(x) = x 2 +6x a. Find the derivative f (x) of f(x). b. Find the point on the graph of f(x) where the tangent line to the curve is horizontal. c. What is the rate of change of f(x) at the point found in part (b). 3

4 Example Find the equation of a line that is tangent to the graph of f(x) = x 3 +2 and parallel to the line 3x y 4 = 0. For find a tangent line so we need two things: a) point b) slope Find the slope from the line y = 3x 4: the SLOPE is m = 3. Now we find the point(s) (x and y values) on f(x) = x that have slope 3 so we need to solve for the derivative (SLOPE of the tangent line) and set it equal to 3. 3 = lim h 0 f(x+h) f(x) h 4

5 4.2 Basic Rules of Differentiation NOTATION proof. The following are all different ways of writing the derivative of y = f(x): f dy (x), dx, d[f(x)] dx, y We have four basic rules for taking derivatives. These four rules are provided without Rule 1: Derivative of a constant d dx (c) = 0 Example Find the derivatives of the following functions. a) y = 7 b) y = π Rule 2: The Power Rule if n is any real number, then d dx (xn ) = nx n 1 Example Find the derivatives 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 and f(x) is a differentiable function then d dx (cf(x)) = c d dx (f(x)) 5

6 Example Find the derivatives of the following functions. a) y = 7x 3 b) y = 4 x 3 c) y = 1 4x 3 Rule 4: The Sum and Difference Rule If f(x) and g(x) are differentiable functions then d dx [f(x)±g(x)] = d dx [f(x)]± d dx [g(x)] Example Find the derivatives of the following functions. a) y = 4x 5 +3x 4 8x 2 +x+e 2 b) y = 7x (5x) 7 6

7 c) f(t) = 2t 2 t 3 Example Find 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 ): y y 1 = m(x x 1 ) 1. f(x) = 2, (2, 2) 2. f(x) = 4 x, (16,2) 3. f(x) = 8 x 3, (2,0) 7

8 4. f(x) = 2 9x 2, ( 1, 2 ) 9 Example Let f(x) = x 3 4x 2. Find the point(s) on the graph of f where the tangent line is horizontal. Example The supply function for a certain make of satellite radio is given by p = f(x) = x 5/4 +10 where x is the quantity supplied and p is the unit price in dollars. a. Find f (x). b. What is the rate of change of the unit price if the quantity supplied is 10,000 satelite radios? 8

9 4.3 The Product and Quotient Rules The Product Rule If f(x) and g(x) are differentiable functions then d d [f(x)g(x)] = f(x) dx dx [g(x)]+g(x) d dx [f(x)] It is often easier to think of this rule in terms of words: The first times the derivative of the second plus the second times the derivative 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) and g(x) are differentiable functions then d dx [ ] f(x) = g(x) g(x) d dx [f(x)] f(x) d dx [g(x)] [f(x)] 2 NOTE: There is a negative sign in this equation and yes, it does matter which way you subtract. 9

10 This one is also easier to remember in words: The bottom times the derivative of the top minus the top times the derivative of the bottom all divided by the bottom squared. Example f(t) = t2 t 3 +1 = top bottom Example g(x) = 3 x x+1 = top bottom Rewrite this first. You need exponents that are numbers. Example h(t) = 1 5t 2 10

11 Example f(x) = ( 4+ 1 )(2x 1x ) x 2 Example f(s) = s+3 3 s 2 Example g(t) = (t+1)(t2 +1) t 2 11

12 Higher Order Derivatives We can find the derivative of the derivative, this is known as the second derivative. This tells us the slope of the derivative. Similarly we can find the derivative of the second derivative, this is called the third derivative. Notation: We write the higher order derivatives as follows: Original Function: y = f(x) First derivative: y, f (x), dy dx, d dx [f(x)] Second derivative: y, f (x), d 2 y dx 2, d 2 dx 2 [f(x)] Third derivative: y, f (x), d 3 y dx 3, d 3 dx 3 [f(x)] Fourth derivative: y (4), f (4) (x), d 4 y dx 4, d 4 dx 4 [f(x)] Example Find the first, second and third derivatives of the functions. 1. f(x) = x 5 x 4 +3x 3 +3x 2 x 9 2. f(x) = 1 x 12

13 4.4 The Chain Rule The chain rule is used for composition of functions: y = (f g)(x) = f(g(x)) There are two methods that can be used to solve for the derivative of a composition of functions. Both methods are called the Chain Rule and are essentially the same using different notation. Method 1: If we rewrite the functions as y = f(u) and u = g(x) where both y and u dy du are differentiable functions then we can take the derivatives separately: and If we du dx multiply them together we will get an expression for dy dx : dy dx = dy du dudx Example Differentiate y = (x 5 8x) 2 using Method 1. Indicate which function is y and which is u. Here we have a composition of functions. We can think of this as y = f(u) = u 2 and u = g(x) = x 5 8x If we take the composition of these two functions we get: 13

14 Q. Why use the chain rule? In the previous example we could have easily multiplied out the function and taken a derivative. 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 complicated. Example Differentiate y = (x 5 8x) 25 using Method 1. Indicate which function is y and which is u. Method 2: If y = (f g)(x) = f(g(x)) then y = f (g(x)) g (x) Both method 1 and method 2 are known as the chain rule. The second method may be easier. In the second method it is often easiest to think of these functions as the Outside function and the Inside function. The outside function is f(u) and the inside function is u = g(x). It is often easiest to remember the chain rule in words: Take the derivative of the outside function (leave the inside function alone) then multiply by the derivative of the inside function. 14

15 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 derivatives of a variable raised to an exponent: y = x n, but with the chain rule we can take the derivative of a function raised to an exponent. The General Power Rule Given the function then or y = [u(x)] n dy dx = n[u(x)]n 1 du dx y = n u n 1 u Example y = 1 x+1 15

16 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 product rule with the chain rule: Example y = x 4 x4 +4 Here we will use the quotient rule with the chain rule: 16

17 ( t 2 Example y = t 3 +2 ) 3 Example Find the third derivative of g(t) = (3t 2 1) 5. 17

18 4.5 Implicit Differentiation Explicit Functions: Definition 4.1. An explicit function is a function in which one variable is defined only in terms of the other variable. (a) y = x 2 +7 Parabola (b) y = 9 x 2 Top half of the circle centered at the origin with radius 3. (c) y = 9 x 2 Bottom half of the circle centered at the origin with radius 3. Implicit Functions: Definition 4.2. An implicit function is a function in which one variable is not defined only in terms of another variable. (a) y x 2 = 7 Parabola (b) y 2 +x 2 = 9 Circle centered at the origin with radius 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 and (b) can be solved for either the top or the bottom of the circle. (c) can not be solved explicitly for y in terms of x. Implicit Differentiation: We have seen how to differentiate functions of the form y = f(x). We also want to be able to differentiate functions that either can t be written explicitly in terms of x or the resulting function is too complicated to deal with. To do this we use implicit differentiation. Key to implicit differentiation (chain rule): d dx [yn ] = ny n 1dy dx and d dx [xn ] = nx n 1dx dx = nxn 1 18

19 Steps to find dy dx of implicit functions involving the variables x and y. 1. Treat y as a differentiable function of x. 2. Differentiate both sides of the equation with respect to x using all the rules we have d previously used. i.e. Apply the operator to every term and use the product, dx quotient and chain rules. 3. Solve for dy dx as if it were a variable. (a) Simplify each side of the equal sign. (b) Bring all dy terms to one side of the equal sign and all non-dy dx dx other. (c) Factor out dy dx. (d) Divide to isolate dy dx. terms to the Example Find dy dx for x3 x 2 xy = 4 (a) by solving the equation explicitly in terms of y and (b) by implicit differentiation. 19

20 Example Find dy dx for y2 +x 2 = 9 Example Find dy dx for x2 y 2 +y 3 x = x 3 y 3 20