Selamat HUT RI ke 63

Merdeka!!!
Hari ini seluruh rakyat indonesia merayakan kemerdekaan RI yang ke 63 tahun. Tidak terkecuali di SMAN 1 UNGARAN, kita juga ngerayain lho kemerdekaan. Pagi-pagi kita sudah berangkat dengan penuh semangat menuju lapangan upacara.
Kita semua sadar banget akan arti penting kemerdekaan bagi bangsa kita. Makanya, kita melaksanakan upacara ini dengan penuh hikmad. Apalagi ketika Pasukan Paskibra mulai beraksi.

derivatives

DERIVATIVES

Definition:

If f(x) is a function, then the derivative of f(x) is:

Example:

Find the derivative of f(x) = x² + 1

Answer:

Notations of derivative: .

THE FORMULAS OF DERIVATIVE

1. F(x) = c, c R then .

Example:

f(x) = 5 then f’(x) = 0.

2. F(x) = ax + b, then .

Example:

F(x) = 3x then f’(x) = 3.

3. F(x) = axⁿ then f’(x) = n.ax^n-1.

Example:

Find the derivatives of the following functions:

a. F(x) = x² b. f(x) = x³ c. f(x) = 4x⁶

Answer:

a. F(x) = x², then f’(x) = 2x.

b. f(x) = x³, then f’(x) = 3x².

c. f(x) = 4x⁶, then f’(x) = 24x⁵.

4. F(x) = u(x) v(x) then f’(x) = u’(x) v’(x).

Example:

Find the derivatives of the following functions:

a. F(x) = 3x² + 6x + 2 b. f(x) = x³ – 5x² + c. f(x) = (3x+1)² d. f(x) =

Answer:

a. F(x) = 3x² + 6x + 2 c. f(x) = (3x+1)² = 9x² + 6x +1

F’(x) = 6x + 6. f’(x) = 18x + 6.

b. f(x) = x³ – 5x² + d. f(x) =

= x³ – 5x² + 8x^-1 = =

f’(x) = 3x² – 10x – 8x^-2 f’(x) =

= 3x² – 10x - . = = .

5. y = u.v then y’ = u’.v + v’.u. (u and v is a function)

Example:

Find the derivatives of:

a. f(x) = (x² – 1)(3x + 2) b. f(x) = (x⁴ + 5)(x² + 3)

Answer:

a. f(x) = (x² – 1)(3x + 2) b. f(x) = (x⁴ + 5)(x² + 3)

u = x² – 1 → u’ = 2x u = x⁴ + 5 → u’ = 4x³

v = 3x + 2 → v’ = 3 v = x² + 3 → v’ = 2x

f’(x) = u’.v + v’.u f’(x) = u’.v + v’.u

= 2x.( 3x + 2) + 3.( x² – 1) = 4x³.( x² + 3) + 2x.(x⁴ + 5)

= 6x² + 4x + 3x² – 3. = 4x⁵ + 12x³ + 2x⁵ + 10x

= 6x⁵ + 12x³ + 10x.

6. then .

Example:

Find the derivative of .

Answer:

.

7. F(x) = {u(x)}ⁿ then f’(x) = n.{u(x)}^n-1.u’(x)

Example:

Find the derivatives of :

a. F(x) = (3x + 1)⁵ b. f(x) =

Answer:

a. F(x) = (3x + 1)⁵ b. f(x) = =

F’(x) = 5.(3x + 1)⁴.3 f’(x) =

= 15(3x + 1)⁴. = (x+2).

TANGENT EQUATION

The tangent equation through point (x₁,y₁) with the slope (gradient) m:

y – y₁ = m(x – x₁)

where m =f’(a) = (the derivative of y for x = a)

Examples:

1. Find the tangent equation on curve y = x³ + 2x through point (1,3)

2. Find the tangent equation on curve y = x² – x + 3 through absis point 2

3. Find the tangent equation on curve y = x² + 4 with the slope 8.

Answer:

1. y = x³ + 2x 3. y = x² + 4 m = 8

y’ = 3x² + 2 y’ = 2x ↔ 8 = 2x

(1,3) then x = 1 → m = y’ = 3.(1)² + 2 = 5. x = 4

Tangent equation through (1,3) with m = 5 x = 4 → y = (4)² + 4

y – y₁ = m(x – x₁) = 20.

y – 3 = 5(x – 1) (4,20)

y – 3 = 5x – 5 T.e. through (4,20) with m = 8

y = 5x – 2. y – y₁ = m(x – x₁)

So, the tangent equation is y = 5x – 2. y – 20 = 8(x – 4)

2. y = x² – x + 3 y – 20 = 8x – 32

y’ = 2x – 1 y = 8x – 12.

x = 2 → m = y’ = 2.(2) – 1 = 3. So, the tang. Eq. is y = 8x – 12.

X = 2 → y = (2)² – 2 + 3 = 5. (2,5)

Tangent equation through (2,5) with m = 3

y – y₁ = m(x – x₁)

y – 5 = 3(x – 2)

y – 5 = 3x – 6

y = 3x – 1.

So, the tangent equation is y = 3x – 1.

Notes:

Parallel lines → m₁ = m₂

Perpendicular → m₁.m₂ = -1

Examples:

1. Find the tangent equation of touching curve y = x² – x + 1 which parallel to y – 5x + 2 = 0.

2. Find the tangent equation of touching curve y = x² + 2x + 4 which perpendicular to x – 4y + 3 = 0.

Answer:

1. line y – 5x + 2 = 0 m₁ = 5 x = 3 y = (3)² – 3 + 1 = 7

parallel m₁=m₂=5 tangent point (3,7)

y = x² – x + 1 tangent equation through (3,7) with m = 5

y’ = 2x – 1 y – y₁ = m(x – x₁)

5 = 2x – 1 (m₂=5) y – 7 = 5(x – 3)

2x = 6 y – 7 = 5x – 15

x = 3 y = 5x – 8

2. line x – 4y + 3 = 0 m₁ = x = -3 y = (-3)² + 2(-3) + 4 = 7

perpendicular m₁.m₂ = -1 tangent point (-3,7)

. m₂ = -1 tangent equation through (-3,7) with m = -4

m₂ = -4 y – y₁ = m(x – x₁)

curve y = x² + 2x + 4 y – 7 = -4(x + 3)

y’ = 2x + 2 y – 7 = -4x -12

-4 = 2x + 2 (m₂ = -4) y = -4x -5

2x = -6 So, the tangent equation is y = -4x -5

x = -3

THE FUNCTION INCREASES AND THE FUNCTION DECREASES

F(x) increases if f’(x) > 0.

F(x) decreases if f’(x) <>

Examples:

1. Find the interval when f(x) = x³ + 3x² + 5 decreases!

2. Find the interval when f(x) = 5 + 15x + 9x² + x³ increases!

Answer:

1. F’(x) = 3x² + 6x <>

x² + 2x <>

x(x + 2) <>

x = 0 / x = -2

+++ - - - +++

-2 0

-2 < style=""> So, the decreasing interval is -2 <>

2. F’(x) = 0 + 15 + 18x + 3x² > 0

3x² + 18x + 15 > 0

x² + 6x + 5 > 0

(x + 5)(x + 1) > 0

x = -5 / x = -1

+++ - - - +++

-5 -1

x < -5 x > -1

So, the decreasing interval x < -5 or x > -1

STATIONER POINT

F(x) stationer if f’(x) = 0.

Types of stationer points:

1.

[a,f’(a)]

Maximum point Point of inflexion (titik belok)

f’(a) is called maximum value.

a

[a,f’(a)] is called maximum point.

2. Minimum point

f’’(x) <>

f’(a) is called minimum value.

[a,f’(a)] is called minimum point.

[a,f’’(a)] is called inflexion point.