Lesson 6: The Derivative

Transcription

1 Lesson 6: Te Derivative Def. A difference quotient for a function as te form f(x + ) f(x) (x + ) x f(x + x) f(x) (x + x) x f(a + ) f(a) (a + ) a Notice tat a difference quotient always as te form of cange in y divided by cange in x wic sould make us tink of te slope of a line. In fact, tat s exactly wat it is, te slope of a line passing troug 2 points. Def. Te derivative of a function is te slope of te tangent line to te function. Looking at te top grap, te dased line is te tangent line to f(x) at te point. If we want to find te slope of tis line, we need 2 points on te line, but we only ave te 1 point. However, we can estimate te tangent line by looking at a point close to te point. Looking at te bottom grap, we ave picked suc a point. Te slope of tis line as te form of te difference quotient defined above. slope = f(x + ) f(x) (x + ) x = f(x + ) f(x) If goes to 0, ten we re moving towards te tangent line at te actual point were we want to find te slope. Tis leads to anoter definition of te derivative. 1

2 Def. Te derivative of f(x) is given by f (x) f(x + ) f(x) Wen we use te definition of te derivative to find a derivative, we are using te limit process to find te derivative. Tere are different notations for te derivative. Most commonly, we see f (x), dy dx, y, and d dx f(x). Ex.1 Find te derivative of f(x) = 3 x. We need to find We ave... f (x) f(x + ) = 3 (x + ) = 3 x f(x + ) f(x) f(x + ) f(x) = (3 x ) (3 x) = 3 x 3 + x = f(x+) f(x) = = 1 f (x) f(x + ) f(x) ( 1) = 1 Since we re taking te limit as goes to 0 of -1, and -1 doesn t depend on, te limit is just 1. Be careful wit parenteses! You need to distribute all negatives! 2

3 Ex.2 Find te derivative of f(x) = 1 x 1. We re going to go troug te limit process of finding te derivative. f (x) f(x + ) f(x) 1 1 (x+) 1 x 1 x 1 x+ 1 (x+ 1)(x 1) (x 1)(x+ 1) (x 1) (x+ 1) (x+ 1)(x 1) x 1 x + 1 (x + 1)(x 1) 1 = 1 (x 1) 2 (x + 1)(x 1) 1 1 (x + 1)(x 1) Looking at te simplifications above, we first find a common denominator for te numerator to combine te 2 fractions. Ten we simplify until we get rid of te 1. Finally, we can take te limit wit 0 to get rid of and get te derivative. 3

4 Ex.3 Find te equation of te tangent line to te grap of f(x) = 2 x at x = 1. Finding te equation of a tangent line to f(x) at x = c is a very important concept tat will appear trougout tis course. I break te process into 3 steps. Note tat te equation of a tangent line will ALWAYS ave te form y = mx + b. Step 1: Find te slope of te tangent line to f(x) at x = c, i.e. m = f (c). Step 2: Find a point were te tangent line touces te grap at x = c, i.e. te point (c, f(c)). Step 3: Use te slope and point to find te equation of te tangent line. Now let s work troug tis example. Step 1: Find m = f (1). f (x) f(x + ) f(x) (2 x + ) (2 x) 2 x x x + + x At tis point, we still need to cancel te from te denominator so tat te limit is someting we can evaluate. Unfortunately, te in te numerator is trapped inside a square root. To liberate from tis annoying prison, we need to rationalize te expression in te numerator. Remember from pre-calculus tat rationalizing someting uses te difference of 2 squares: a 2 b 2 = (a b)(a + b). We need to recognize if we ave (a b) or (a + b) ten multiply by te conjugate (te factor wit te opposite sign). Ten we ave te difference of 2 squares and we can square te terms, so te variables or numbers are no longer trapped under te square root. In tis instance, we ave (a + b) wit a = x + and b = x. Tis means we are going to multiply and divide te fraction by te conjugate wic is (a b) = x + x. (We could look at x x + and use different definitions of a and b, but let s continue wit te order as written above.) 4

5 f (x) ( x + + x x + x x + x x + ) 2 ( x) 2 ( x + x) (x + ) (x) ( x + x) ( x + x) 1 ( x + x) = 1 2 x Now tat we ave f (x) = 1 2 x, we can find te slope m = f (1) = 1 2. Step 2: Find te point were te tangent line and function touc, i.e. (1, f(1)). f(1) = 2 1 = 1 Step 3: Find te equation of te tangent line y = mx + b. We ave 2 metods tat we usually use to find te equation of line. We can use point-slope form or we can find te y-intercept. y-intercept Metod: Plug te slope m = 1 2 mx + b and solve for b. and point (x, y) = (1, 1) into y = y = mx + b 1 = ( 1 2 )(1) + b 1 = b = b 3 2 = b Now we plug m and b into y = mx + b and we ave te tangent line y = 1 2 x Point-slope metod Start wit y y 1 = m(x x 1 ). Plug in m = 1 2 and (x 1, y 1 ) = (1, 1) ten solve for y. 5

6 y y 1 = m(x x 1 ) y 1 = 1 (x 1) 2 y 1 = 1 2 x y = 1 2 x y = 1 2 x You can see tat bot metods give te tangent line y = 1 2 x Here are a few examples to try on your own: Ex.4 Find te derivative of f(x) = 2 x. Answer: f (x) = x Ex.5 Find te equation of te tangent line to te grap of f(x) = x at x = 0. Answer: Slope is m = f (0) = 0. Equation of te tangent line is y = 4. 6