# DIFFERENTIAL IMMERSION

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### DIFFERENTIALorDERIVATIVE :

DIFFERENTIALorDERIVATIVE

### A. Derivative of Algebra function :

A. Derivative of Algebra function The 1st derivative of y = f(x) can be written y’ = f’(x) = = and is defined :

### Example : :

Example : Use formula before to find f’(x) f(x) = c f(x) = x f(x) = x2 f(x) = f(x) =

c.

d.

e.

### Slide 8:

Note: f(x) = C ? f’(x) = 0 f(x) = x ? f’(x) = 1 f(x) =x2 ? f’(x) = 2x f(x) = ? f’(x) = f(x) = 1/x = x-1 ? f’(x) = f(x) = axn ? f’(x) = n axn-1 So if f(x) = axn ? f’(x) = n.a.xn-1

### Slide 9:

Example: Find f’(x) f(x) = f(x) = f(x) = f(x) =

c.

d.

### Slide 13:

2. If Find : a. f’(x) b. if f’(x) = 0 find x c. if f’(x) < 0 find x d. if f’(x) = 0 find x e. if f’(x) = 3 find x Answer f’(x)= 2x2 –x-3 b.

c. d.

e.

### B. DERIVATIVE FORMULA :

B. DERIVATIVE FORMULA f(x) = C ? f’(x) = 0 f(x) = axn ? f’(x) = a.nxn-1 f(x) = U.V ? f’(x) =U’.V + V’.U f(x) = f(x) = U ± V ? f’(x) = U’± V’ f(x) = a Un ? f’(x) = a.n.U’.Un-1

### Slide 17:

Example Find f’(x)

c. d.

e.

f.

### C. Derivative of Trigonometric Function :

C. Derivative of Trigonometric Function Use the formula To find: y = sin x y = sin (ax±b) y = cos x y = cos (ax±b) y = tan x y = cotan x (cosx/sinx)

### CONCLUSION :

CONCLUSION Based on the problem above we can make conclusion: f(x) = sin x ? f’(x) = cos x f(x) = sin (ax±b) ? f’(x) = a cos (ax±b) f(x) = cos x ? f’(x) = -sinx f(x) = cos (ax±b) ? f’(x) = -a sin (ax±b) f(x) = tan x ? f’(x) = sec2 x f(x) = tan (ax±b) ? f’(x) = a sec2 (ax±b) f(x) = cotan x ? f’(x) = -cosec2 x f(x) = cotan (ax±b) ? f’(x) = -acosec2 (ax±b)

### Note! :

Note! f(x) = sinn u ? f’(x) = n.u’ cos u sinn-1.u f(x) = cosn u ? f’(x) = -n.u’ sin u cosn-1.u f(x) = tann u ? f’(x) = n.u sec2 u tan n-1.u f(x) = cotann u ? f’(x) = - n.u’cosec2.u.cotann-1.u

### EXAMPLE :

EXAMPLE Find f’(x) f(x) = 4sinx – 2cosx + tanx f’(x) = 4cosx + 2sinx + sec2x 2. f(x) = 3sin(-2x2) + 5cos(-x3+1) – 3 cotan4x f’(x) = -12x cos(-2x2) + 15x2 sin(-x3+1) + 12 cosec24x 3. f(x) = -3cos42x f’(x) = 24 sin2x cos32x = 12. 2 sin2x cos2x cos22x = 12 sin4x cos22x

### EXAMPLE :

EXAMPLE 4. 5. 6. 7. 8. 9. 10. 11.

### NOTE :

NOTE f(x) = sec x =

### D. The Equation of Tangent Line of Curve y=f(x) :

D. The Equation of Tangent Line of Curve y=f(x) P(a,f(a))and Q((a+h),f(a+h))?

### Slide 32:

If Q point move approach P point then h=0 Mpq= Conclusion: Slope of y=f(x) mx=a is m=f’(a) The equation of tangent line of y=f(x) in (x1,y1) y-y1=m(x-x1)

### Example: :

Example: Determine the equation tangent line of the both below: a. b. c. m=8 d. Parralel to the line e. Perpendicular the line With intersect point in y axis With intersect point in x axis

### Example: :

Example: 2. The equation tangent line in (2,0) has slope=1. The equation tangent line in (0,6) parallel 5x+y=4. Find a, b, c, and d!

### E. UP function and DOWN function :

E. UP function and DOWN function At g1 the curve y=f(x) its function is down so, m=f’(x)<0 At g2 the curve y=f(x) its function is up so, m=f’(x)>0 At g3 the curve y=f(x) reach the culmination point so m=f’(x)<0

### Conclusion :

Conclusion Y=f(x) is called up function if f’(x)>0 Y=f(x) is called down function if f’(x)<0 Y=f(x) is never up and never down function if f’(x)=0 (condition stationer value) Y=f(x) is called never up if f’(x)=0 Y=f(x) is called never down if f’(x)=0

### EXAMPLE: :

EXAMPLE: Determine interval x in order to f(x) up and f(x) down a. b. c. d. e.

### EXAMPLE: :

EXAMPLE: 2. Prove that a. Always up b. Always down c. Always down d. Always up Remember x2 always =0 -x2 always = x2+c always >0 -x2+c always <0

ANSWER 2. a. f’(x)=x2+2 so f(x) always up (prove) b. f’(x) So f(x) always down(prove)

### F. STATIONARY VALUE, STATIONARY POINT AND THE KINDS OF STATIONARY POINT :

F. STATIONARY VALUE, STATIONARY POINT AND THE KINDS OF STATIONARY POINT

### A. Stationary Value :

y=f(x) (a, f(a))= stationary point f(a)= stationary value x=a A. Stationary Value

### B. The kinds of stationary point :

(i) y=f(x) (a, f(a))= minimum point f(a)= min stationary point x=a B. The kinds of stationary point

### B. The kinds of stationary point :

(a,f(a)) = maximum point y = f(x) (a, f(a))= minimum point f(a)= min stationary point x=a B. The kinds of stationary point

### B. The kinds of stationary point :

(i) (a, f(a))= turning point f(a)= turning point x=a B. The kinds of stationary point

### B. The kinds of stationary point :

(i) (a, f(a))= turning point y = f(x) x=a B. The kinds of stationary point

### Example :

Determine stationary value, stationary point, and kinds of it: f(x)= x³- 6x² - 15x + 2 f(x)= -x³ +5x² + 8x +2 f(x)= 4x³ - 3x4 f(x)= x5-10x³ + 40x f(x)= x4- 4x³ f(x)= 2 sinx-1 0=x= 360 f(x)= sin 2x 0=x= 360 f(x)= c0s x+7 0=x= 360 Example

Answer 1. F(-1) = (-1)3-6(-1)2-15(-1)+2 = -1-6+15+2 = 10 ? Max Stationary value (-1,10) ? Max point F(5) = (5)2-6.52-15.5+2 = 125-150-75+2 = -98 ? Min Stationary value (5,-98) ? Min point

### G. Drawing Graph of function :

Determine the intersection point y axis so that condition x=0 Determine the intersection point x axis so that condition y=0 Find the kind of stationary point f¹(x)=0 Please take helping point, look the value of x which made stationary point ? surrounding x= of stationary point G. Drawing Graph of function

### Slide 53:

Draw the graph of the function below y = x4-2x2 y = x4-4x3 -1<x<3 y = sin x – cos x o0<x<3600 y = x3-3x y = 3x5-5x3 y = 6x-x3 -3<x<3 F(x) = -2 sin 2x

### Slide 54:

Answer a. F(-1) = (-1)4-2(-1)2 = 1-2 = -1 (-1,-1) = min point F(0) = 0 (0,0) max point F(1) = 1-2 = -1 (-1,-1) min point

### Maximum and Minimum value at close interval a<x<b :

Maximum and Minimum value at close interval a<x<b a b d e f g c at interval a<x<c ? maximum value = b minimum value = c at interval b<x<e ? maximum value = f(e) minimum value = f(d) at interval a<x<g ? maximum value = f(f) minimum value = f(d)

### Slide 57:

Conclusion:From the result above, actually maximum stationary value and minimum stationary value aren’t certain as maximum value and minimum value. So that to find max & min value can be done by: Find stationary value (x must be element interval) Find the corner point of interval Find from the result no 1 and 2, then choose the max & min value that we need

### Slide 58:

Determine max & min value from that function below f(x) = x2-8x 0<x<5 f(x) = x3-3x 0<x<3 f(x) = x4-2x2-x -2<x<3 f(x) = (x+1)3.(x-2) <x<1 f(x) = -2sin x 0<x<

### Slide 59:

Answer (i) f(4) = 42-8.4 = 16-32 = -16 NU f(0)=0 f(5)=-15 Max value = 0 Min value = -16

### Using Max and Min Value in Daily Life :

Using Max and Min Value in Daily Life Example A piece of carton in the square will be made a box without cover. If dimension of carton 16 cm, determine dimension of a box in order that max value and the max value if the corner of carton cut x cm. As like no 1 but in the form of rectangle with the length 24 cm and width 9 cm The missile track can be seen with the formula h(t)=400t-5t2. When does the missile reach the max high and find the max high. Determine the max value of right cylinder inscribe the sphere with the radius 6 cm.

### Slide 61:

The area of rectangle 121 cm2. In order to it’s perimeter max, determine the dimension of rectangle and perimeter of it. A box without cover base of it in the form of square will be made from a piece of carton with the volume that we want is 32 dm3. Determine the dimension of a box in order to the material that use minimum and the area of it. Multiplying of the both positive number is 64. Determine the both number in order that has min sum and find min sum of that number.

### Using the Second Derivative to Determine the Kinds of Stationary Point :

Using the Second Derivative to Determine the Kinds of Stationary Point From To determine the kind of stationary value and point by the second derivative in x=a, can be done by: if f”(a) < 0 ? (a,f(a)) max stat point if f”(a) > 0 ? (a,f(a)) min stat point if f”(a) = 0 ? stat value and stat point can’t be determined by 2nd derivative.

### Slide 64:

Example F(x) = x3-6x2 F(x) = 3x2-x3 F(x) = 4x3-x4 F(x) = 3x5-5x3 F(x) = 2x2-1 F(x) = -x2+3 F(x) = 2x5-1 F(x) = 2x4

### Slide 65:

Answer a. X=0 ? f”(0)=0-12=-12<0 f(0) = 0-0 = 0 max stat value (0,0) max stat point X=4 ? f”(4)=24-12=12>0 f(4)=64-96=-32 min stat value (4,-32)

### The Distance, Velocity and Acceleration :

The Distance, Velocity and Acceleration If s = f(t) then, the formula of

### Slide 67:

Example The material move in the straight line with track formula s=3-6t+2t3. determine The formula of velocity and acceleration The velocity at t = 2 The acceleration at t = 3 The velocity in time a = 0

### Slide 68:

As like no 1, s = t3-2t2+4t Find average velocity if t = 2s until t = 3s Determine the formula of velocity and acceleration As like no 1 s = t3-6t2+12t+1. how much the time is needed in order to that acc = 4

### Slide 69:

As like no 1, , t = 0determine the distance that velocity in the time acc = 0

### Slide 70:

Answer a. The formula of velocity & acceleration v = s” a = 12t s’ = 6t2-6 b. The velocity at t = 2 v = 2 ? v = 24-16 = 18 m/s2 c. The acceleration at t = 3 a(3) = 36 m/s2 d. The velocity in time a = 0 a = 0 ? v = 0 v(0) = -6 