Maths Level 4 Module 6

Mathematics NQF Level 4 *see terms and conditions Downloaded from hlayiso.com Mathematics NQF Level 4

Functions and algebra Topic 2 Downloaded from hlayiso.com Mathematics NQF Level 4

The instantaneous rate of change of a variable Module 6 Downloaded from hlayiso.com Mathematics NQF Level 4

VIDEO Differentiation from first principles Click above to play this video. Downloaded from hlayiso.com Unit 6.1

Rules for differentiating functions Downloaded from hlayiso.com Unit 6.2

Rules for differentiating functions Downloaded from hlayiso.com Unit 6.2

Learning activity 6.3 Downloaded from hlayiso.com Mathematics NQF Level 4

dy Determine if: dx 1. y = 4 x 3 + 2 x − 6 dy d 3 d d [Sum/difference = 4 x +2 x− 6 dx dx dx dx rule] = 4.3 x 2 + 2.1 − 0 = 12 x 2 + 2 Downloaded from hlayiso.com Learning activity 6.3

dy Determine if: dx 2. y = a + e + π + cos 14° + 2 x dy d d d d d = a + e + π + cos14° + 2 x dx dx dx dx dx dx = 0+0+0+0+2 =2 [Sum/difference rule] Downloaded from hlayiso.com Learning activity 6.3

dy Determine if: dx 1 5. y= 2e + 3ln x − 5 + 2 x x dy d x d d d −2 = 2 e + 3 ln x − 5 + x dx dx dx dx dx 1 [Sum/difference = 2e + 3 ⋅ − 0 + −2 x −3 x x rule] 3 2 = 2e + − 3 x x x Downloaded from hlayiso.com Learning activity 6.3

dy Determine if: dx [Sum/difference sin x + 2 cos x − 5 tan x 6. y = rule] dy d d d = sin x + 2 cos x − 5 tan x dx dx dx dx 2 = cos x + 2(− sin x) − 5(sec x) 2 =cos x − 2sin x − 5sec x Downloaded from hlayiso.com Learning activity 6.3

Learning activity 6.4 Downloaded from hlayiso.com Mathematics NQF Level 4

Differentiate with respect to x 2 5) y = 13 x − 5 x + 8 VIDEO Click above to play this video. Downloaded from hlayiso.com Learning activity 6.4

Learning activity 6.5 Downloaded from hlayiso.com Mathematics NQF Level 4

Differentiate with respect to x by using the product rule: = 9. y 3ln 4 x ⋅ sin 2 x Let u 3ln 4 x = and v sin 2 x du dv du = u +v dx dx dx d d 3ln 4 x sin 2 x + sin 2 x 3ln 4 x dx dx Downloaded from hlayiso.com Learning activity 6.5

Differentiate with respect to x by using the product rule: 9. = y 3ln 4 x ⋅ sin 2 x Continued 4 [Chain rule] = 3ln 4 x ⋅ cos 2 x ⋅ 2 + sin 2 x ⋅ 3 ⋅ 4x 3 = 6 ln 4 x ⋅ cos 2 x + sin 2 x x Downloaded from hlayiso.com Learning activity 6.5

Learning activity 6.6 Downloaded from hlayiso.com Mathematics NQF Level 4

Differentiate with respect to x by using the quotient rule: x 4) y = 2 2 Let = u x and = v 2x − 4 2x − 4 du dv v −u dy = dx 2 dx [Quotient rule] dx v 1 2 d d 2 2x − 4 x − x (2 x − 4) 2 = dx dx 2 2x − 4 Downloaded from hlayiso.com Learning activity 6.6

Differentiate with respect to x by using the quotient rule: x 4) y = Continued 2 2x − 4 1 2 1 2 − 2 x − 4 − x ⋅ (2 x − 4) ⋅ 4 x2 2 [Chain rule] = 2x2 − 4  2   2 2x  1 = 2x − 4 −  1  2 2x − 4 2 2   (2 x − 4)  Downloaded from hlayiso.com Learning activity 6.6

Differentiate with respect to x by using the quotient rule: x 4) y = Continued 2x2 − 4 2 1 2x 1 − 3 (2 x 2 − 4) 2 (2 x 2 − 4) 2 2 2 (2 x − 4) − 2 x = 3 (2 x 2 − 4) 2 −4 = 3 (2 x 2 −Downloaded 4) 2 from hlayiso.com Learning activity 6.6

Learning activity 6.8 Downloaded from hlayiso.com Mathematics NQF Level 4

1. An object is projected vertically upwards (against gravity).Its displacement (s) in metres during a period of time (t) in seconds is given by s = 32t− 5t2.Calculate the: 1.1) Displacement after 2s 2 s 32t − 5t = ∴ f ( t ) =32t − 5t 2 (2) 32(2) − 5(2) 2 f= = 64 − 20 = 44m Downloaded from hlayiso.com Learning activity 6.8

1. 1.2) Velocity after 2 s f ( t=) 32t − 5t 2 = f '(t= v ) 32 − 10t ∴ f '(2) = 32 − 10(2) = 12 m/s Downloaded from hlayiso.com Learning activity 6.8

1. 1.3) Time taken to reach a maximum height For max. height: f '(t ) = 0 ∴ 32 − 10t = 0 ∴ t =3, 2s 1.4) Maximum height f ( t=) 32t − 5t 2 2 f (3, = 2) 32(3, 2) − 5(3, 2) = 51, 2 m Downloaded from hlayiso.com Learning activity 6.8

1. 1.5) Displacement after 4 s f ( t=) 32t − 5t 2 (4) 32(4) − 5(4) 2 f= = 48 m 1.6) Distance after 4 s s =51, 2 + (51, 2 − 48) = 51, 2 + 3, 2 = 54, 4 m Downloaded from hlayiso.com Learning activity 6.8

2. Air is let out of a large, round balloon at a rate of 0,4 m3/s.At what rate will the radius decrease if the radius is 3 m? VIDEO Click above to play this video. Downloaded from hlayiso.com Learning activity 6.8

Summative assessment Module 6 Downloaded from hlayiso.com Mathematics NQF Level 4

1. Find the derivatives of the following functions from first principles. 1.1) f ( x = ) 3x + 2 2 2 f ( x + h) = 3( x + h) + 2 2 2 = 3( x + 2 xh + h ) + 2 2 2 = 3 x + 6 xh + 3h + 2 f ( x + h) − f ( x ) f '( x) = lim h →0 h Downloaded from hlayiso.com Summative assessment – Module 6

1. Find the derivatives of the following functions from first principles. 1.1) f ( x = ) 3x + 2 2 Continued 2 2 2 3 x + 6 xh + 3h + 2 − (3 x + 2) = lim h →0 h 6 xh + 3h 2 = lim h →0 h h(6 x + 3h) = lim h →0 h = 6x Downloaded from hlayiso.com Summative assessment – Module 6

1. Find the derivatives of the following functions from first principles. 1.2) f ( x ) = 2 x 3 3 f ( x + h) = 2( x + h) = 2( x + h)( x 2 + 2hx + h 2 ) 3 2 2 3 =2 x + 6hx + 6h x + 2h f ( x + h) − f ( x ) f '( x) = lim h →0 h Downloaded from hlayiso.com Summative assessment – Module 6

1. Find the derivatives of the following functions from first principles. 1.2) f ( x ) = 2 x 3 2 x 2 + 6 x 2 h + 6h 2 x + 2h 3 − 2 x 3 = lim h →0 h 2 2 h(6 x + 6hx + 2h ) = lim h →0 h 2 = 6x Downloaded from hlayiso.com Summative assessment – Module 6

2. Use the chain rule to differentiate y = 3cos2x with respect to x Let u = 2 x ⇒ y = 3cos u du =2 and dy = −3sin u dx du dy dy du = ⋅ = −3sin u ⋅ 2 dx du dx = −6sin 2x Downloaded from hlayiso.com Summative assessment – Module 6

3. Differentiate with respect to x 4x 3.1) 3e d4x = 3e ⋅ 4 x [Chain rule] dx 4x = 3e ⋅ 4 [Exponential rule] 4x = 12e Downloaded from hlayiso.com Summative assessment – Module 6

3. Differentiate with respect to x 3.2) y = 2 ln 3 x dy d [Chain rule] = 2 ln 3 x dx dx 1 d = 2⋅ 3x [Log rule] 3 x dx 2 = ⋅3 3x 2 = x Downloaded from hlayiso.com Summative assessment – Module 6

dy 4. Determine if: dx ( 4x + 4) 2 7 = 4.1) y dy 6 d 2 7(4 x + 4) = 2 (4 x + 4) [Chain rule] dx dx = 7(4 x 2 + 4)6 (8 x) 56 x(4 x 2 + 4)6 Downloaded from hlayiso.com Summative assessment – Module 6

dy 4. Determine if: dx 4.2) y = 4 tan 3 x 2d = 4sec 3 x 3 x [Chain rule] dx 2 4sec 3 x ⋅ 3 2 = 12sec 3x Downloaded from hlayiso.com Summative assessment – Module 6

dy 1  1 5. Determine if y= x  + 4  − + 4 x . dx x  x 1  1 y= x  + 4  − + 4 x . x  x 1 dy d 1 = (1 + 4 x − + 4 x 2 ) [Sum/difference rule] dx dx x 1 −2 1 − = 0 + 4(1) − (− x ) + 4 ⋅ x 2 2 1 2 =4 + 2 + x x Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx 6.1) = y x 2 x3 + 1 1 3 =y x(2 x + 1) 2 1 1 dy d d [Product = (2 x + 1) ⋅ x + x (2 x 3 + 1) 3 2 2 dx dx dx rule] 3 1  1 3 − 1 d 3  = (2 x + 1) ⋅1 + x  (2 x + 1) 2 2 (2 x + 1)  2 dx  Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx = 6.1) y x 2 x3 + 1 Continued 3  6 x 2  = (2 x + 1) + x    2 (2 x3 + 1)  6 x3 = 2 x3 + 1 + 2 2x2 + 1 Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx = 6.2) y 2e ⋅ cos 2x −x dy d d − x [Product 2e −x cos 2 x + cos 2 x 2e dx dx dx rule] = 2e (− sin 2 x)2 + cos 2 x ⋅ 2e (−1) −x −x −4e = −x sin 2 x − 2e − x cos 2 x Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: NOTE: When you dx are familiar with the technique, 2x some of the steps 6.3) y = can left out for 1− x simplicity Let u = 2 x and v = 1− x du dv v −u dy [Quotient rule] = dx 2 dx dx v 1 d d 1 − x 2 x − 2 x (1 − x) 2 dy dx dx = dx 1− x Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx 2x 6.3) y = Continued 1− x 1 1 − 2 1 − x − 2 x ⋅ (1 − x) 2 (−1) = 2 1− x  x  1 =  2 1− x + × 1− x  1− x  Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx 2x 6.3) y = Continued 1− x 2 − 2x + x = 3 (1 − x) 2 2− x = 3 (1 − x) 2 Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx x 6.4) y = 3x e d 3x d 3x e x−x e [Quotient rule] = dx dx 6x e 3x 3x e − x ⋅e ⋅3 = 6x e Downloaded from hlayiso.com Summative assessment – Module 6

dy 6. Determine if: dx x 6.4) y = 3x Continued e e3 x ⋅ (1 − 3 x) = e6 x 1 − 3x = 3x e Downloaded from hlayiso.com Summative assessment – Module 6

7. Calculate the equation of the tangent to y= 3 + 1 at x = 2 x 3 f ( x)= +1 x 3 ∴ Tangent at At x = 2 : f (2)= + 1 = 2,5 2 (2; 2,5) Find the gradient of the tangent: ′ m = f ( x) = −3 x −2 3 ∴m = − 4 Downloaded from hlayiso.com Summative assessment – Module 6

7. Calculate the equation of the tangent to y= 3 + 1 at x = 2 x Find equation of tangent: y − y1 = m( x − x1 ) [Gradient-point formula] −3 y − 2,5= ( x − 2) 4 5 −3 3 y− = x+ 2 4 2 −3 =y x+4 4 Downloaded from hlayiso.com Summative assessment – Module 6

8. Sand falls onto a conical pile at a rate of 3 m3/min. and the radius of the conical base is always equal to one half of its altitude. How fast is the altitude of the pile increasing when it is 4,5 m high? 1 r= h [Given] 2 dV [Given] = 3m3 / min dt Figure 6.15 Downloaded from hlayiso.com Summative assessment – Module 6

8. Continued 1 = π ⋅ r 2h V [Volume of cone] 3 2 1 1  =π ⋅  h  ⋅ h 3 2  1 ∴V = π h3 12 dV 1 2 ∴ = πh dh 4 Figure 6.15 Downloaded from hlayiso.com Summative assessment – Module 6

8. Continued dV dV dh = ⋅ [Chain rule] dt dh dt 1 2 dh ∴3 = π h . 4 dt dh 4 ∴= ⋅ 3 [h = 4,5m] dt π (4,5) 2 dh ∴ 0,189 m/min = dt Figure 6.15 ∴ The altitude of the pile is increasing by 0,189m/min Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.1) Determine the intercepts with the axes. x-intercept ⇒ y = 0 y -intercept ⇒ x =0 ∴ x3 − 2 x 2 − 3x = 0 ∴y =0 ∴ x( x 2 − 2 x − 3) = 0 ∴ x( x − 3)( x + 1) = 0 ∴ x =0 or x =3 or x =−1 Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.2) Find the turning point(s) and indicate the minimum and maximum. 3 2 f ( x) =x − 2 x − 3 x f ′( x) = 3 x 2 − 4 x − 3 f ''( x= ) 6x − 4 Turning point: f ′( x) = 0 3x 2 − 4 x − 3 =0 Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.2) Find the turning point(s) and indicate the minimum and maximum. (Continued) 4 ± 16 − 4(3)(−3) x= 6 4 ± 7, 211 = 6 ∴ x =1,869 or x =−0,535 Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.2) Find the turning point(s) and indicate the minimum and maximum. (Continued) Substitute the x-values to find corresponding y -values: f (1,869) =(1,869)3 − 2(1,869) 2 − 3(1,869) = 6,529 − 6,986 − 5, 607 = −6, 064 (1,869; − 6, 064) f ''(1,869) = 6(1,869) − 4 > 0 ∴ minimum point Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.2) Find the turning point(s) and indicate the minimum and maximum. (Continued) f (−0,535) =(−0,535)3 − 2(−0,535) 2 − 3(−0,535) = −0,153 − 0,572 + 1, 605 = 0,88 (−0,535; 0,88) f ''(−0,535) = 6(−0,535) − 4 < 0 ∴ maximum point Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.3) Calculate the point(s) of inflection. Points of inflection: f ′′( x) = 0 6x − 4 =0 x = 0, 667 Substitute the x-value to find corresponding y -value: f (0, 667) =(0, 667)3 − 2(0, 667) 2 − 3(0, 667) = 0, 297 − 0,890 − 2, 001 = −2,594 (0, 667; − 2,594) Downloaded from hlayiso.com Summative assessment – Module 6

3 2 9. Given y =x − 2 x − 3 x 9.4) Sketch the graph. Figure 6.16 Downloaded from hlayiso.com Summative assessment – Module 6

Mathematics NQF Level 2 Downloaded from hlayiso.com Mathematics NQF Level 4