Using the definition of the derivative of a function is quite tedious. f (x + h) f (x)

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1 Derivative Rules Using te efinition of te erivative of a function is quite teious. Let s prove some sortcuts tat we can use. Recall tat te efinition of erivative is: Given any number x for wic te limit f (x) 0 f (x + ) f (x) exists, we assign to x te number f (x). We now give some basic erivative rules for fining erivatives witout aving to use te limit efinition irectly. AMAT 217 (University of Calgary) Fall / 19

2 Constant Function: (c) = 0. x Proof: Let f (x) = c be a constant function. By te efinition of erivative: f (x) 0 f (x + ) f (x) 0 c c = 0 Wy can 0 above be replace by 0? Recall: Wit limits lim x a f (x), we only care about wat appens for values of x arbitrarily close to a but not equal to a. Tus, we are only consiering values of close to 0, but not equal to 0. AMAT 217 (University of Calgary) Fall / 19

3 Te Power Rule: If n is a positive integer, ten x (x n ) = nx n 1. Proof: We use te formula: x n a n = (x a)(x n 1 + x n 2 a + + xa n 2 + a n 1 ) wic can be verifie by multiplying out te rigt sie. Let f (x) = x n be a power function for some positive integer n. Ten at any number a we ave: f (a) x a f (x) f (a) x a x n a n x a x a x a (x a)(x n 1 + x n 2 a + + xa n 2 + a n 1 ) x a x a (x n 1 + x n 2 a + + xa n 2 + a n 1 ) = na n 1. Note: Tis rule ols for every real n. AMAT 217 (University of Calgary) Fall / 19

4 Te Constant Multiple Rule: If c is a constant an f is a ifferentiable function: Proof: For convenience let g(x) = cf (x). Ten: g (x) 0 g(x + ) g(x) cf (x + ) cf (x) 0 0 c [ ] f (x + ) f (x) = c lim 0 f (x + ) f (x) = cf (x) were c can be move in front of te limit by Limit Rule 3. x [cf (x)] = c x f (x). AMAT 217 (University of Calgary) Fall / 19

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6 Te Sum/Difference Rule: If f an g are ifferentiable: x [f (x) ± g(x)] = x f (x) ± x g(x). Proof: For convenience let r(x) = f (x) ± g(x). Ten: r (x) 0 r(x + ) r(x) [f (x + ) ± g(x + )] [f (x) ± g(x)] 0 [ f (x + ) f (x) ± 0 ] g(x + ) g(x) f (x + ) f (x) g(x + ) g(x) ± lim 0 0 = f (x) ± g (x) were te limit can be istribute across ± by Limit Rule 2. AMAT 217 (University of Calgary) Fall / 19

7 Te Prouct Rule: If f an g are bot ifferentiable: x [f (x) g(x)] = f (x) x [g(x)] + g(x) [f (x)] x Proof: For convenience let r(x) = f (x) g(x). As in te previous proof, we want to separate te functions f an g. Te trick is to a an subtract f (x + )g(x) in te numerator. r f (x + )g(x + ) f (x)g(x) (x) 0 f (x + )g(x + ) f(x + )g(x) + f(x + )g(x) f (x)g(x) 0 [ ] g(x + ) g(x) f (x + ) f (x) f (x + ) + g(x) 0 g(x + ) g(x) f (x + ) f (x) f (x + ) lim + lim g(x) lim = f (x)g (x) + g(x)f (x) AMAT 217 (University of Calgary) Fall / 19

8 Te Quotient Rule: If f an g are bot ifferentiable: x [ ] f (x) = g(x) x [f (x)] f (x) x [g(x)] g(x) [g(x)] 2 Proof: Te proof is similar to te previous proof but te trick is to a an subtract te term f (x)g(x) in te numerator. We omit te etails. AMAT 217 (University of Calgary) Fall / 19

9 Sine Function: (sin x) = cos x Proof: Let f (x) = sin x. Using te efinition of erivative we ave: f (x) 0 f (x + ) f (x) 0 sin(x + ) sin x 0 [sin x cos + cos x sin ] sin x 0 [sin x cos sin x] + [cos x sin ] cos 1 sin sin x lim + lim cos x lim = sin x 0 + cos x 1 = cos x sin x Tis follows from te formulas lim x 0 x be prove using te squeeze teorem. cos x 1 = 1 an lim = 0 wic can x 0 x AMAT 217 (University of Calgary) Fall / 19

10 A similar proof works for te cosine function: (cos x) = sin x. x Using quotient rule we get formulas for te oter trigonometric ratios. Derivatives of Trigonometric Functions (sin x) = cos x x (cos x) = sin x x x (tan x) = sec2 x (csc x) = csc x cot x x (sec x) = sec x tan x x x (cot x) = csc2 x AMAT 217 (University of Calgary) Fall / 19

11 Summary of erivative rules (so far) Derivative Rules 1. (c) = 0 2. (x n ) = nx n 1 3. (cf ) = cf 4. (f ± g) = f ± g 5. (fg) = f g + fg 6. ( ) f = f g fg g g 2 7. (sin(x)) = cos(x) 8. (cos(x)) = sin(x) Number 2 is te Power Rule. Number 3 is te Constant Multiple Rule. Number 4 is te Sum/Difference Rule. Number 5 is te Prouct Rule. Number 6 is te Quotient Rule. AMAT 217 (University of Calgary) Fall / 19

12 Example Fin te erivative of f (x) = 17. Solution: Te erivative of a constant is 0, so we ave f (x) = 0. Example Fin te erivative of g(x) = x 4. Solution: By te Power Rule, we bring te exponent own an ecrease te exponent by one: g (x) = 4x 3. AMAT 217 (University of Calgary) Fall / 19

13 Example Fin te erivative of (x) = 17 + x 4. Solution: We ave tat (x) = f (x) + g(x), were f (x) = 17 an g(x) = x 4. From te first two problems we know: Example f (x) = 0 an g (x) = 4x 3 By te Sum/Difference Rule, we ave Fin te erivative of F (x) = 2 (17 + x 4 ). (x) = f (x) + g (x) = 0 + 4x 3 = 4x 3 Solution: By te Constant Multiple Rule, we can take te erivative of 17 + x 4 (wic we i in te previous question) an ten multiply by 2, so, F (x) = 2(4x 3 ) = 8x 3. AMAT 217 (University of Calgary) Fall / 19

14 Example Fin te vertex of te parabola y = x 2 4x + 5 (witout completing te square). Solution: Te vertex of a parabola as a orizontal tangent line wit slope zero. Terefore, at te vertex te erivative must equal zero! For our example, y = x 2 4x + 5, we get y x = 2x 4 = 0 2x = 4 x = 2 Ten plugging x = 2 into te equation to get y: y = 2 2 4(2) + 5 y = y = 1 Terefore, te vertex is at (2, 1). AMAT 217 (University of Calgary) Fall / 19

15 Example (Fall Final Exam - Q1) Fin te erivative of f (x) = (x 2 2) sin(x) + 2x cos(x). Solution: f (x) = ( [2x] [sin x] + [x 2 2] [cos x] ) + ([2] [cos x] + [2x] [ sin x]) Prouct Rule twice = 2x sin x + x 2 cos x 2 cos x + 2 cos x 2x sin x Expaning = x 2 cos x Cancelling AMAT 217 (University of Calgary) Fall / 19

16 Example (Fall Miterm Exam - Q11) { 1 4x if x < 1 Fin f ( 1), if it exists, given tat f (x) = x if x 1 Solution: First, we nee to verify f (x) is continuous at x = 1 (omework). In tis case it is continuous (if it was not continuous ten we woul automatically conclue tat f ( 1) oes not exist). From te left of x = 1 we ave: f ( 1) = x ( 1 4x) x= 1 = 4 From te rigt of x = 1 we ave: f +( 1) = x ( x ) x= 1 = 4x 3 x= 1 = 4( 1) 3 = 4 Terefore, f ( 1) = 4. AMAT 217 (University of Calgary) Fall / 19

17 Example Given F (x) = Solution: By te Quotient Rule we ave: x 1 + (x), (0) = 3 an (x) is ifferentiable, fin F (0). F (x) = Tus, wen x = 0 we ave: ( ) f = f g fg g g 2 wit f (x) = x an g(x) = 1 + (x), x (x) (1 + (x)) (x) x (1 + (x)) (1 + (x)) 2 = 1(1 + (x)) (x)( (x)) (1 + (x)) 2 = 1 + (x) x (x) (1 + (x)) 2 F (0) = 1 + (0) 0 (1 + (0)) 2 = (1 + 3) 2 = 1 4. AMAT 217 (University of Calgary) Fall / 19

18 Example Fin te erivative of (x) = (3x 1)(2x + 3). Solution: One way to o tis question is to expan te expression. Alternatively, we use te Prouct Rule (fg) = f g + fg wit f (x) = 3x 1 an g(x) = 2x + 3. Note tat f (x) = 3 an g (x) = 2, so, (x) = (3) (2x + 3) + (3x 1) (2) = 6x x 2 = 12x + 7. AMAT 217 (University of Calgary) Fall / 19

19 Example Fin te erivative of (x) = 3x 1 2x + 3. Solution: By te Quotient Rule ( ) f = f g fg g g 2 wit f (x) = 3x 1 an g(x) = 2x + 3, we ave: (x) = = = x (3x 1) (2x + 3) (3x 1) x (2x + 3) (2x + 3) 2 3(2x + 3) (3x 1)(2) (2x + 3) 2 11 (2x + 3) 2 AMAT 217 (University of Calgary) Fall / 19