Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Te Derivative as a Function DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. f (a) f(a+) f(a) REMARK: An equivalent way of stating te definition of te derivative is f (a) OTHER NOTATIONS: f (x) y dy dx df dx d dx f(x) Df(x) D xf(x) EXAMPLE: If f(x) 3x, find f (x). Solution : We ave f (a) f(a+) f(a) Solution 2: We ave f (a) ( ) 3(a+) (3a ) 3a+3 3a+ (3x ) (3a ) 3x 3a+ 3 3 3 3x 3a 3() 3 3 EXAMPLE: If f(x) x 2, find f (x).
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: If f(x) x 2, find f (x). Solution : We ave f (a) f(a+) f(a) (a+) 2 a 2 a 2 +2a+ 2 a 2 2a+ 2 (2a+) (2a+) 2a+0 2a or f (a) f(a+) f(a) (a+) 2 a 2 (a+ a)(a++a) (2a+) (2a+) 2a+0 2a Solution 2: We ave f (a) x 2 a 2 ()(x+a) (x+a) a+a 2a 2
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: If f(x) x, find f (x). Solution : We ave f (a) x a ( x) 2 ( a) 2 ()( x+ a) Solution : We ave a+ a 2 a f (a) Solution 2: We ave f f(a+) f(a) (a) a+) 2 ( a) 2 ( x a x a ( x a)( x+ a) ( a++ a) a++ a Solution 2 : We ave f f(a+) f(a) (a) a+ a ( x a)( x+ a) ()( x+ a) ()( x+ a) x+ a a+ a ( a++ a) a+0+ a a+ a ( a+) 2 ( a) 2 a++ a x a ( x) 2 ( a) 2 x+ a a+ a 2 a ( a+ a)( a++ a) ( a++ a) a+ a 2 a a+ a a+0+ a ( a++ a) a+ a a+ a a+ a ( a+ a)( a++ a) a+ a 2 a EXAMPLE: If f(x) 2x 3 x, find f (x). Solution: We ave f (a) 2x 3 2a 3 x+a (2x 3 x) (2a 3 a) 2(x 3 a 3 ) () 2x 3 x 2a 3 +a ( ) 2(x 2 +xa+a 2 ) 2()(x 2 +xa+a 2 ) () () ( ) 2(x 2 +xa+a 2 ) 2(a 2 +a a+a 2 ) 2(3a 2 ) 6a 2 3
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: If f(x) x+3 4 x, find f (x). Solution : We ave f (a) x+3 4 x a+3 4 a x+3 a+3 (4 x)(4 a) 4 x 4 a (4 x)(4 a) () (4 x)(4 a) (x+3)(4 a) (a+3)(4 x) ()(4 x)(4 a) 4x xa+2 3a 4a+ax 2+3x ()(4 x)(4 a) 7() ()(4 x)(4 a) ( x+3 4 x a+3 4 a ) (4 x)(4 a) () (4 x)(4 a) (4x xa+2 3a) (4a ax+2 3x) ()(4 x)(4 a) Solution 2: We ave (a+)+3 f f(a+) f(a) 4 (a+) a+3 4 a (a) ( a++3 4 a a+3 ) (4 a )(4 a) 4 a (4 a )(4 a) 7x 7a ()(4 x)(4 a) 7 (4 x)(4 a) 7 (4 a)(4 a) 7 (4 a) 2 a++3 a+3 (4 a )(4 a) 4 a 4 a (4 a )(4 a) (4 a )(4 a) (a++3)(4 a) (a+3)(4 a ) (4 a )(4 a) a++3 4 a a+3 4 a (4a a 2 +4 a+2 3a) (4a a 2 a+2 3a 3) (4 a )(4 a) 7 (4 a )(4 a) 7 (4 a )(4 a) 7 (4 a 0)(4 a) 7 (4 a) 2 EXAMPLE: Use te grap of a function f to sketc te grap of te derivative f. 4
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: Use te grap of a function f to sketc te grap of te derivative f. Solution: We ave DEFINITION: A function f is differentiable at a if f (a) exists. It is differentiable on an open interval (a,b) [or (a, ) or (,a) or (, )] if it is differentiable at every number in te interval. EXAMPLES:. A polynomial, sin x, cos x are differentiable everywere. 2. Te function f(x) x in differentiable on (,0) and on (0, ). EXAMPLE: Sow tat f(x) x is not differentiable at x 0.
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: Sow tat f(x) x is not differentiable at x 0. Solution: Note tat { x if x < 0 x x if x 0 Terefore on te one and we ave f(0+) f(0) 0+ 0 lim On te oter and, f(0+) f(0) lim + + 0+ 0 Since f(0+) f(0) lim f(0+) f(0) it follows tat lim is not differentiable at x 0. + + lim + f(0+) f(0) does not exist. Terefore f (0) does not exist, so f(x) x How can a function fail to be differentiable? Tere are tree main instances wen f is not differentiable at a:. Wen te grap of f as a corner or kink in it. 2. Wen f is not continuous at a. 3. Wen te grap of f as a vertical tangent line at x a; tat is, wen f is continuous at a and lim f (x). 6
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: Matc te grap of eac function in (a)-(d) wit te grap of its derivative in I-IV. Give reasons for your coice. THEOREM: If f is differentiable at a, ten f is continuous at a. REMARK: Te opposite statement is not true! Tat is, tere are functions tat are continuous at a certain point but not differentiable. For example, f(x) x. EXAMPLE: Te grap is given. State, wit reasons, te numbers at wic f is not differentiable. I II III IV 7
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: Te grap is given. State, wit reasons, te numbers at wic f is not differentiable. I II III IV Solution: (I) f is not differentiable at x 4 (corner) and x 0 (discontinuity) (II) f is not differentiable at x 0 (discontinuity) and x 3 (corner) (III) f is not differentiable at x (vertical tangent) and x 4 (corner) (IV) f is not differentiable at x (discontinuity) and x 2 (corner) Higer derivatives If f is a differentiable function, ten f is also a function. So, f may ave a derivative of its own, denoted by (f ) f. Tis new function f is called te second derivative of f. OTHER NOTATIONS: f (x) y d dx ( ) dy d2 y dx dx 2 EXAMPLE: We already know tat if f(x) 2x 3 x, ten f (x) 6x 2. Te second derivative is f (a) f (a+) f (a) 6a 2 +2a+6 2 6a 2 + (2a+6) 6(a+) 2 (6a 2 ) (2a+6) 2a 2a+6 2 8
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Te tird derivative f is te derivative of te second derivative f (f ). OTHER NOTATIONS: f (x) y d dx ( ) d 2 y dx 2 d3 y dx 3 Similarly, te fourt derivative f is te derivative of te tird derivative f (f ). And so on. REMARK: Te fourt derivative f is usually denoted by f (4). In general, f (n) (x) y (n) dn y dx n Application In general, we can interpret a second derivative as a rate of cange of a rate of cange. Te most familiar example of tis is acceleration, wic we define as follows. If s s(t) is te position function of an object tat moves in a straigt line, we know tat its first derivative represents te velocity v(t) of te object as a function of time: v(t) s (t) ds dt Te instantaneous rate of cange of velocity wit respect to time is called te acceleration a(t) of te object. Tus te acceleration function is te derivative of te velocity function and is terefore te second derivative of te position function: a(t) v (t) s (t) or, in Leibniz notation, a dv dt d2 s dt 2 9
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Appendix EXAMPLE: If f(x) 3 2x 7x 2, find f (x). Solution : We ave f (a) f(a+) f(a) Solution 2: We ave f (a) ) ) (3 2(a+) 7(a+) 2 (3 2a 7a 2 3 2a 2 7(a 2 +2a+ 2 ) 3+2a+7a 2 3 2a 2 7a 2 4a 7 2 3+2a+7a 2 2 4a 7 2 ( 2 4a 7) ( 2 4a 7) 2 4a 7 0 2 4a (3 2x 7x 2 ) (3 2a 7a 2 ) 3 2x 7x 2 3+2a+7a 2 2x 7x 2 +2a+7a 2 2x+2a 7x 2 +7a 2 2() 7(x 2 a 2 ) 2() 7()(x+a) ( ) () 2 7(x+a) ( ) 2 7(x+a) 2 7(a+a) 2 7(2a) 2 4a 0
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: If f(x) x+9, find f (x). Solution : We ave In sort, f (a) ( x+9 a+9 x+9 a+9)( x+9+ a+9) ()( x+9+ a+9) ( x+9) 2 ( a+9) 2 ()( x+9+ a+9) (x+9) (a+9) ()( x+9+ a+9) x+9 a 9 ()( x+9+ a+9) x a ()( x+9+ a+9) () ()( x+9+ a+9) x+9+ a+9 a+9+ a+9 2 a+9 f (a) x+9 a+9 ( x+9 a+9)( x+9+ a+9) ()( x+9+ a+9) ( x+9) 2 ( a+9) 2 ()( x+9+ a+9) (x+9) (a+9) ()( x+9+ a+9) () ()( x+9+ a+9) x+9+ a+9 2 a+9
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Solution 2: We ave f (a) f(a+) f(a) In sort, ( (a+)+9 a+9 (a+)+9 a+9)( (a+)+9+ a+9) ( (a+)+9+ a+9) ( (a+)+9) 2 ( a+9) 2 ( (a+)+9+ a+9) ((a+)+9) (a+9) ( (a+)+9+ a+9) a++9 a 9 ( (a+)+9+ a+9) ( (a+)+9+ a+9) (a+)+9+ a+9 (a+0)+9+ a+9 a+9+ a+9 2 a+9 f (a) (a+)+9 a+9 ( (a+)+9 a+9)( (a+)+9+ a+9) ( (a+)+9+ a+9) ( (a+)+9) 2 ( a+9) 2 ( (a+)+9+ a+9) ( (a+)+9+ a+9) ((a+)+9) (a+9) ( (a+)+9+ a+9) (a+)+9+ a+9 2 a+9 2
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka EXAMPLE: If f(x) 2 3x, find f (x). Solution: We ave In sort, f (a) ( 2 3x 2 3a 2 3x 2 3a)( 2 3x+ 2 3a) ()( 2 3x+ 2 3a) ( 2 3x) 2 ( 2 3a) 2 ()( 2 3x+ 2 3a) (2 3x) (2 3a) ()( 2 3x+ 2 3a) 2 3x 2+3a ()( 2 3x+ 2 3a) 3x+3a ()( 2 3x+ 2 3a) 3() ()( 2 3x+ 2 3a) 3 2 3x+ 2 3a 3 2 3a+ 2 3a 3 2 2 3a f (a) 2 3x 2 3a ( 2 3x 2 3a)( 2 3x+ 2 3a) ()( 2 3x+ 2 3a) ( 2 3x) 2 ( 2 3a) 2 ()( 2 3x+ 2 3a) (2 3x) (2 3a) ()( 2 3x+ 2 3a) 3() ()( 2 3x+ 2 3a) 3 3 2 3x+ 2 3a 2 2 3a 3