S5_Unit_3_Outcome_2 - Higher Unit 3 Higher Outcome 2...

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Unformatted text preview: Higher Unit 3 Higher Outcome 2 www.mathsrevision.com Differentiation The Chain Rule Further Differentiation Trig Functions Further Integration Integrating Trig Functions www.mathsrevision.com T h e C h a in R ule fo r Diffe re ntia ting Higher Outcome 2 To differentiate composite functions (such as functions with brackets in them) we can use: www.mathsrevision.com dy dy du = dx du dx Example Differentiate y = ( x 2 + 3)8 then y = u 8 dy = 8u 7 du Let u = ( x 2 + 3) du = 2x dx dy = dx dy du du dx = 8u 7 2 x = 8( x 2 + 3)7 .2 x dy = 16 x( x 2 + 3)7 dx Higher You have 1 minute to 1. Differentiate outside the bracket. come up with the rule. Outcome 2 2. Keep the bracket the same. Good News ! 3. Differentiate inside the bracket. The Chain Rule for Differentiating There is an easier way. www.mathsrevision.com h '( x) = g '( f ( x)) f '( x) Differentiate h( x) = ( x 2 + 3)8 h '( x) = 16 x( x 2 + 3) 7 Let f ( x) = ( x 2 + 3) then g ( x) = x8 g ( f ( x )) = ( x 2 + 3)8 g '( f ( x )) = 8( x 2 + 3) 7 h '( x ) = 16 x( x 2 + 3)7 f '( x ) = 2 x 1. Differentiate outside the bracket. Higher 2. Keep the bracket the same. Outcome 2 Example 3. Differentiate inside the bracket. Find f '( x) when f ( x) = The Chain Rule for Differentiating www.mathsrevision.com ( x 2 + 3) -4 5 4 f ( x) = 5( x + 3) 2 f '( x) = -20( x + 3) 2 -5 f '( x) = -20( x + 3) 2 -5 2x -40 x f '( x) = 2 ( x + 3)5 You are expected to do the chain rule all at once 1. Differentiate outside the bracket. Higher Outcome 2 Example 3. Differentiate inside the bracket. 2. Keep the bracket the same. The Chain Rule for Differentiating www.mathsrevision.com Find dy 1 when y = dx 2x -1 dy = (2 x - 1) dx - 1 2 3 dy 1 = - ( 2 x -1) 2 2 dx 2 dy = - dx 1 (2 x - 1) 3 2 The Chain Rule for Differentiating Higher Example 3 Outcome 2 2 5 If f ( x) = (3 x - 2 x + 1) , find the value of f '(0) www.mathsrevision.com Let f '( x) = 5(3 x - 2 x + 1) (9 x - 4 x) 3 2 4 2 Let f '(0) = 5(3 - 2 + 1) 4 0) 0 0 (9 0 3 2 4 2 f '(0) = 0 The Chain Rule for Differentiating Functions Higher Example Outcome 2 1 where x = 3 2x -1 Find the equation of the tangent to the curve y = www.mathsrevision.com The slope of the tangent is given by the derivative of the equation. y= 1 2x -1 Rearrange: y = ( 2 x - 1) -1 Use the chain rule: dy -2 = - 1( 2 x - 1) 2= dx 1 y= 5 and dy -2 = dx 25 -2 ( 2 x - 1) 2 Where x = 3: -2 1 We need the equation of the line through 3, with slope 5 25 The Chain Rule for Differentiating Functions Higher Outcome 2 1 -2 y = ( x - 3) - 5 25 -2 1 ( x - 3) + = 25 5 -1 5 (2 x - 6) + 25 25 Remember y b = m(x a) www.mathsrevision.com y= y= -1 (2 x - 1) 25 Is the required equation The Chain Rule for Differentiating Functions Higher Outcome 2 Example In a small factory the cost, C, in pounds of assembling x components in a month is given by: 1000 C ( x) = x + 40 , x 2 www.mathsrevision.com x 0 Calculate the minimum cost of production in any month, and the corresponding number of components that are required to be assembled. Minimum occurs where dC = 0, so differentiate dx 2 Rearrange 25 C ( x) = + = 40 x x 25 1600 + x x 2 The Chain Rule for Differentiating Functions Higher Outcome 2 2 dC d 25 = + 1600 x = x dx dx d 25 1600 + x dx x 25 25 3200 + 1 - 2 x x x 2 www.mathsrevision.com Using chain rule 25 25 = 1600 2 + 1 - 2 x = x x or 25 1 = - 2 0 x dC 25 = 0 where + 0 x = dx x i.e. where x 2 = -25 or x 2 = 25 We must have x>0 No real values satisfy x 2 = -25 We must have x = 5 items assembled per month The Chain Rule for Differentiating Functions Higher Outcome 2 Is x = 5 a minimum in the (complicated) graph? www.mathsrevision.com dC 25 25 = 3200 + 1 - 2 = 0 when x = 5 x dx x x Is this a minimum? Consider the sign of dC dx when x > 5 and x < 5 For x < 5 we have (+ve)(+ve)(ve) = (ve) For x = 5 we have (+ve)(+ve)(0) = 0 For x > 5 we have (+ve)(+ve)(+ve) = (+ve) Therefore x = 5 is a minimum x = 5 The Chain Rule for Differentiating Functions Higher Outcome 2 The cost of production: www.mathsrevision.com 25 C ( x) = 1600 + , x x 25 = 1600 + 5 5 2 2 x=5 = 1600 ( 10 ) 2 = 160,000 Expensive components? Aeroplane parts maybe ? Calculus Revision Differentiate ( x + 2)5 5( x + 2) 4 1 5( x + 2) 4 Chain rule Simplify Back Quit Next Calculus Revision Differentiate (5 x - 1)3 3(5 x - 1) 2 5 15(5 x - 1) 2 Chain Rule Simplify Back Quit Next Calculus Revision Differentiate (5 x 2 - 3x + 2) 4 4(5 x - 3x + 2) ( 10 x - 3) 2 3 Chain Rule Back Quit Next Calculus Revision Differentiate (7 x - 1) 5 2 Chain Rule 5 (7 x - 1) 2 3 2 7 3 2 Simplify 35 (7 x - 1) 2 Back Quit Next Calculus Revision Differentiate (2 x + 5) - 1 2 Chain Rule 1 - (2 x + 5) 2 - 3 2 2 Simplify -(2 x + 5) - 3 2 Back Quit Next Calculus Revision Differentiate Straight line form 3x - 1 ( 3x - 1) 1 2 1 2 Chain Rule ( 3x - 1) 3 2 - 1 2 1 2 3 Simplify ( 3x - 1) - Back Quit Next Calculus Revision Differentiate f ( x) = (8 - x ) 1 2 1 3 2 Chain Rule f ( x) = (8 - x 3 ) 3 - 2 - 1 2 ( -3 x 2 ) - 1 2 Simplify f ( x) = x 2 (8 - x 3 ) Back Quit Next Calculus Revision Differentiate f ( x) = ( 5 x - 4 ) 1 2 Chain Rule f ( x) = 1 2 ( 5x - 4) - 1 2 5 Simplify f ( x) = 5 2 ( 5x - 4 ) - 1 2 Back Quit Next Calculus Differentiate (x 5 2 Revision 4 + 3) Straight line form 5( x 2 + 3) -4 -20( x 2 + 3) -5 2 x -40 x( x 2 + 3) -5 Chain Rule Simplify Back Quit Next Calculus Differentiate 1 2x -1 Revision Straight line form (2 x - 1) 1 - (2 x - 1) 2 - 1 2 3 2 Chain Rule - 2 3 2 Simplify -(2 x - 1) - Back Quit Next Trig Function Differentiation Higher Outcome 2 www.mathsrevision.com The Derivatives of sin x & cos x d (sin x) = cos x dx d (cos x) = - sin x dx Trig Function Differentiation Higher Example 2 f ( x) = 2cos x + sin x . 3 Outcome 2 Find f '( x) www.mathsrevision.com d d 2 f '( x) = (2cos x) + sin x dx dx 3 d [ f ( x) + g ( x)] = dx df dg + dx dx =2 d 2 d (cos x) + ( sin x ) dx 3 dx d dg c.g ( x) = c dx dx 2 f '( x) = -2sin x + cos x 3 Trig Function Differentiation Higher Example Differentiate Outcome 2 4 - x 3 cos x with respect to x 3 x Simplify expression where possible www.mathsrevision.com 4 - x cos x = 4 x -3 - cos x x3 3 d (4 x -3 - cos x) = dx = -12 x -4 - (- sin x) = -12 x -4 + sin x x 4 sin x - 12 = x4 d d 4 x -3 - (cos x) dx dx Restore the original form of expression Higher The Chain Rule for Differentiating 1. Differentiate outside the bracket. Trig Functions Outcome 2 2. Keep the bracket the same. 3. Differentiate inside the bracket. www.mathsrevision.com Worked Example: dy = cos dx Differentiate y = sin(4 x) dy = cos 4 x du dy = cos 4 x 4 dx dy = 4cos 4 x dx Higher Example The Chain Rule for Differentiating Trig Functions Outcome 2 Find www.mathsrevision.com dy when y = cos 4 x dx dy = (cos x) 4 dx dy = 4(cos x)3 (-sin x) dx dy = -4sin x cos3 x dx Higher Example The Chain Rule for Differentiating Trig Functions Outcome 2 Find www.mathsrevision.com dy when y = sin x dx y = (sin x) 1 2 1 - dy 1 = (sin x) 2 cos x dx 2 dy cos x = dx 2 sin x Calculus Differentiate 2 2 cos x + sin x 3 2 -2sin x + cos x 3 Revision Back Quit Next Calculus Revision Differentiate -3cos x 3sin x Back Quit Next Calculus Revision Differentiate cos 4 x - 2sin 2 x -4sin 4 x - 4 cos 2 x Back Quit Next Calculus Revision Differentiate cos x -4 cos x sin x 3 4 Back Quit Next Calculus Revision Differentiate sin x (sin x) 1 (sin x) 2 - 1 2 Straight line form Chain Rule 1 2 cos x - 1 2 Simplify 1 cos x(sin x) 2 1 2 cos x sin x Next Back Quit Calculus Revision Differentiate - sin(3 - 2 x) - cos(3 - 2 x) ( -2) 2 cos(3 - 2 x) Chain Rule Simplify Back Quit Next Calculus Differentiate cos 5 x 2 Revision Straight line form 1 cos 5x 2 1 sin 5 x 2 5 sin 5x 2 Chain Rule 5 Simplify Back Quit Next Calculus Revision Differentiate f ( x) = cos(2 x) - 3sin(4 x) f ( x) = -2sin(2 x) - 12 cos(4 x) Back Quit Next Calculus Differentiate y = 2 sin ( Revision x- 6 ) Chain Rule dy = 2 cos x - 6 dx dy = 2 cos x - 6 dx ( ) ) 1 Simplify ( Back Quit Next Calculus Revision Differentiate f ( x) = cos 2 x - sin 2 x f ( x) = -2 cos x sin x - 2 sin x cos x f ( x) = -4 cos x sin x Chain Rule Simplify Back Quit Next You have 1 minute to Integrating Composite Functions come up with the rule. Higher Outcome 2 Harder integration www.mathsrevision.com (3 x - 7) dx = 4 we get (3 x - 7)5 +c 15 (ax + b) n+1 (ax + b) n dx = +c a(n + 1) (3 x - 7) dx = 4 (3 x - 7) 4+1 +c = 3(4 + 1) (3 x - 7)5 +c 15 Integrating Composite Functions 2. Divide by new power. Higher 1. Add one to the power. 3. Compensate for bracket. Outcome 2 www.mathsrevision.com (ax + b) n dx = Example : (ax + b) n+1 +c a (n + 1) dt (4t - 2) 4 Evaluate (4t - 2) -4 dt = (4t + 2) -3 (4t + 2) -3 -3 (4t + 2) -3 4 3) (- (4t - 2) -3 =- +c 12 Integrating Composite Functions Higher 1. Add one to the power. 2. Divide by new power. Outcome 2 Example3. Compensate for bracket. www.mathsrevision.com Evaluate (u - 3) -2 du (u - 3) -1 -1 (u - 3) -1 +c 1 1) (- (u - 3) -1 (u - 3) -1 1 -2 (u - 3) du = +c = - +c -1 (u - 3) You are expected to do the integration rule all at once Integrating Composite Functions Higher Example Outcome 2 Evaluate 2 1 (2 x + 1)3 dx www.mathsrevision.com 2 + 1) 4 1 + 1) 4 (2 (1 x + 1) (2 - 8 8 = 8 1 4 2 = 68 Integrating Composite Functions Higher Example Outcome 2 Evaluate www.mathsrevision.com 4 -1 4 -1 (3 x + 4) dx 4 3 2 (3 x + 4) dx = 3 2 5 x + 4) 2 (3 = 5 3 2 -1 2(3 x + 4) 15 5 2 -1 4 5 5 2 2 2(3 4 + 4) 2(3 ( -1) + 4) = - = 15 15 682 5 1. Add one to the power. 2. Divide by new power. Higher Integrating Functions Outcome 2 Example 3. Compensate for bracket. www.mathsrevision.com Find f ( x) given f '( x) = ( 2 x - 1) 3 and f (1) = 2 f ( x) = So we have: ( 2 x - 1) ( 2 x - 1) dx = 3 Integrating 4 8 +c 16 +c= 8 f (1) = 2 = ( 2 1 - 1) + c = 8 4 1 +c = 8 Giving: 1 15 c = 2- = 8 8 f ( x) = ( 2 x - 1) 8 4 15 + 8 Calculus Revision Integrate (5 + 3 x) 2 dx (5 + 3x) +c 3 3 3 Standard Integral (from Chain Rule) 1 3 (5 + 3 x) 9 +c Back Quit Next Calculus Revision Integrate 1 dx 2 (7 - 3 x) (7 - 3 x) dx -2 Straight line form (7 - 3 x) -1 +c ( -1) ( -3 ) 1 (7 - 3 x) -1 3 +c Back Quit Next Calculus Revision 1 Find 0 (2 x + 3) dx 6 Use standard Integral (from chain rule) x + 3) (2 7 2 0 7 1 + 3) 7 + 3) 7 (2 (0 - 14 14 7 7 5 3 - 14 14 5580.36 - 156.21 5424 (4sf) Back Quit Next Calculus 1 Integrate 0 dx Revision 1 2 ( 3x + 1) 1 Straight line form 2 3 ( 3x + 1) 0 1 ( 3x + 1) 0 - 1 2 dx 2 - ( 3 + 1) 3 3 x + 1) ( 1 3 2 1 2 0 1 2 3 ( 0 + 1) 2 3 2 2 - 4 1 3 3 4 2 - 3 3 Back Quit Next Calculus Revision 2 Find 1 ( 2 x + 1) 3 dx 1 2 Use standard Integral (from chain rule) 2 x + 1) ( 4 2 4 4 + 1) 4 2 + 1) 4 ( ( - 4 2 4 2 4 4 5 3 - 8 8 68 Back Quit Next Calculus 0 Evaluate -3 (2 x + 3) 2 0 Revision Use standard Integral dx (from chain rule) x + 3) (2 3 2 -3 3 (2(0) + 3)3 -3) + 3)3 (2( - 6 6 27 27 + 6 6 54 6 27 - 27 - 6 6 9 Back Quit Next Calculus 1 Evaluate Revision 0 3 2 1 + 3x 0 3 dx 1 0 1 ( 1 + 3x ) dx 2 9 2 9 1 2 ( 1 + 3 x ) 3 3 2 2 9 1 3 2 1 + 3x ) 2 ( 9 0 ( 1 + 3x 0 3 ) 1 ( 2 1 + 3(1) - 9 ) ( 1 + 3(0) 3 ) ( ) 2 4 - 9 3 ( ) 1 3 16 2 - 9 9 14 9 1 Quit 5 9 Back Next Calculus p Revision Find p, given 1 p x dx = 42 2 x = 42 3 1 3 2 p x 1 1 2 dx = 42 2 3 p3 2 3 p - 3 2 2 (1) 3 3 2 = 42 - 2 3 = 42 2 p3 - 2 = 126 p= (2 1 12 3 p3 = 64 p 3 = 642 = 212 ) = 16 Back Quit Next Calculus dy = 6 x 2 - 2 x passes through the point (1, 2). A curve for which dx Express y in terms of x. Revision 6 x3 2 x 2 y= - +c 3 2 Use the point y = 2 x3 - x 2 + c 2 = 2(-1)3 - (-1) 2 + c c=5 y = 2 x3 - x 2 + 5 Back Quit Next Calculus Given the acceleration a is: 1 Revision a = 2(4 - t ) 2 , 0 t 4 a= dv dt If it starts at rest, find an expression for the velocity v where dv = 2(4 - t ) dt 4 v=- 3 4 0=- 3 1 2 v= 2(4 - t ) 3 2 3 2 ( -1) +c 3 4 v = - (4 - t ) 2 + c 3 ( 4-t 3 ) 3 +c Starts at rest, so v = 0, when t = 0 32 0= - +c 3 32 c= 3 4 0=- 3 4 3 ( 4) 3 2 3 +c ( 4) +c v = - (4 - t ) + 32 3 Back Quit Next Integrating Trig Functions Higher Outcome 2 www.mathsrevision.com d (sin x) = cos x dx d (cos x) = - sin x dx Integration is opposite of differentiation cos x dx = sin x + c sin x dx = - cos x + c Worked Example ( 3sin x - 1 cos x ) dx = 2 - 3cos x - 1 sin x + c 2 1. Integrate outside the bracket Higher Integrating Trig Functions 2. Keep the bracket the same 3. Compensate for inside the bracket. Special Trigonometry Integrals are Outcome 2 www.mathsrevision.com cos( ax + b) dx = sin( ax + b) dx = Worked Example 1 sin( ax + b) + c a 1 - cos( ax + b) + c a sin(2 x + 2) dx = 1 - cos(2 x + 2) + c 2 1. Integrate outside the bracket 2. Keep the bracket the same Higher Integrating Trig Functions Outcome 2 Example Compensate for inside the bracket. 3. www.mathsrevision.com Evaluate ( sin 5t + cos(-2t ) ) dt Integrate 1 1 - cos5t + sin 2t + c 5 2 Break up into two easier integrals sin(5 co t )dt + s(-2t ) dt = 1. cos( - ) = cos cos + sin sin Integrate outside the bracket Higher Outcome 2 Example Compensate for inside the bracket. 3. 2. Keep the bracket the same 2 Integrating Trig Functions www.mathsrevision.com Evaluate 3cos - 2 x dx 2 4 Rearrange 2 4 Integrate 2 4 3cos - 2 x dx 2 3sin 2x dx = 2 3 - 2 cos 2 x 4 3 - cos = cos 2 - 2 2 2 4 3 = - ( -1 - 0 ) 2 3 = 2 Integrating Trig Functions (Area) Higher y1 y = - sin x y = cos x Example Outcome 2 0 The diagram shows the graphs of y = sin x and y = cos x -1 x A 2 www.mathsrevision.com a) Find the coordinates of A b) Hence find the shaded area The curves intersect where tan x = -1 x = tan (-1) = -1 cos x = - sin x 90o S T 3 2 3 7 and 4 4 180o A C 0o We want 0 x 3 x= 4 270o Integrating Trig Functions (Area) Higher Outcome 2 The shaded area is given by (top curve - bottom curve) dx 3 4 www.mathsrevision.com Area = 3 4 (cos x - (- sin x)) dx = 3 4 0 (cos x + sin x) dx 0 0 [ sin x - cos x ] 1 1 = + + 1 2 2 3 3 = sin - cos 4 4 - ( sin 0 - cos 0 ) = 1+ 2 Integrating Trig Functions Higher Example b) Hence find cos3 x Outcome 2 a) By writing cos3 x as cos(2 x + x) show that cos3 x = 4cos 3 x - 3cos x www.mathsrevision.com cos(2 x + x) = cos 2 x cos x - sin 2 x sin x Remember cos(x + y) = = (cos 2 x - sin 2 x) cos x - (2sin x cos x)sin x = (cos 2 x - sin 2 x) cos x - 2sin 2 x cos x = cos3 x - 3sin 2 x cos x = cos3 x - 3 ( 1 - cos 2 x ) cos x cos3 x = 4cos3 x - 3cos x Integrating Trig Functions Higher Outcome 2 As shown above www.mathsrevision.com 3 cos3 x = 4cos x - 3cos x 3 1 cos x = (cos3 x + 3cos x) 4 1 (cos3 x + 3cos x) dx = 4 3 1 1 + sin 3 x + 3sin x c 4 3 1 cos x dx = ( sin 3 x + 9sin x ) + c 12 Calculus Revision Find 2 sin t dt 7 2 - cos t + c 7 Back Quit Next Calculus Revision Find 3cos x dx 3sin x + c Back Quit Next Calculus Revision Find -2sin d 2 cos + c Back Quit Next Calculus Revision Integrate (6 x 2 - x + cos x) dx Integrate term by term 6 x3 x 2 - + sin x + c 3 2 Back Quit Next Calculus Revision Find 1 3sin x - cos x dx 2 1 -3cos x - sin x + c 2 Integrate term by term Back Quit Next Calculus Revision Find sin 2 x - cos x - dx 3 4 1 1 - cos 2 x - sin x - 3 2 3 4 + c Back Quit Next Calculus Revision ,1 The curve y = f ( x) passes through the point 12 f ( x) = cos 2 x f ( x) = 1= 1= 1 2 Find f(x) use the given point ,1 12 1 sin 2 x 2 +c 1 sin 2 2 1 2 12 +c c= 3 4 1= 1 sin 2 6 + c sin = 1 6 2 1 sin 2 x 2 +c f ( x) = + 3 4 Back Quit Next Calculus f ( x) = sin(3 x) If Revision passes through the point ( 9 , 1 ) ) 1= 1 - 3 1 2 y = f ( x) express y in terms of x. Use the point f ( x) = 1= 1 - cos(3 x) + c 3 ( 9 , 1 1 - cos 3 3 7 6 + c 9 1 3 1= 1 - cos 3 3 7 6 + c +c =c y = - cos(3x ) + Back Quit Next Calculus Revision dy = 3sin(2 x) passes through the point A curve for which dx 3 y = - cos(2 x) + c Find y in terms of x. 2 ( 5 , 12 3 ) Use the point 3=- 3 2 ( 5 , 12 3 ) 3 = - cos(2 3=- 3 c= 4 3 2 5 )+c 12 5 cos( ) + c 6 3 3 - + c 2 2 3= 3 3 +c 4 4 3 3 3 - =c 4 4 y = - cos(2 x) + 3 2 3 4 Back Quit Next ...
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This note was uploaded on 02/13/2012 for the course MAT 205 math 205 taught by Professor Google during the Spring '10 term at University of Phoenix.

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