Teorema (Penjabaran dari definisi)

1

Turunan Fungsi Aljabar

A. Turunan sebagai Limit Fungsi

∆𝑡 = 𝑡2− 𝑡1 jika dan hanya jika 𝑡2 = ∆𝑡 + 𝑡1

𝑚 =𝑓(𝑡2)−𝑓(𝑡1)

𝑡2−𝑡1 = 𝑓(∆𝑡+𝑡1)−𝑓(𝑡1)

∆𝑡

= 𝑓(∆𝑡+𝑡1)−𝑓(𝑡1)

∆𝑡 , karena melengkung maka

= lim

∆𝑡→0

𝑓(∆𝑡+𝑡1)−𝑓(𝑡1)

∆𝑡

= 𝑓′(𝑡₁)

𝑚 = lim_∆𝑥→0𝑓(∆𝑥+𝑥)−𝑓(𝑥)_∆𝑥 = 𝑓′(𝑥)

𝑓′(𝑥) = lim_∆𝑥→0𝑓(∆𝑥+𝑥)−𝑓(𝑥)_∆𝑥

Definisi (Pengertian)

Misalkan fungsi 𝑓: 𝑆 → 𝑅 dan 𝑆 ⊆ 𝑅, maka fungsi 𝑓 dapat diturunkan di titik 𝑥 jika dan hanya jika lim

∆𝑥→0

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ada. Turunan dapat dinotasikan 𝑓′(𝑥), 𝑦′, 𝑑𝑓(𝑥)

𝑑𝑥 , atau 𝑑𝑦 𝑑𝑥.

Contoh: Jika diketahui 𝑓(𝑥) = 3𝑥2, maka tentukan 𝑓′(𝑥)! Diketahui: 𝑓(𝑥) = 3𝑥2

Ditanya: 𝑓′(𝑥) ? Jawab: 𝑓(𝑥) = 3𝑥2

𝑓(∆𝑥) = 3(∆𝑥)2 𝑓(𝑥 + ∆𝑥) = 3(𝑥 + ∆𝑥)2

2 𝑓′(𝑥) = lim_∆𝑥→0𝑓(∆𝑥 + 𝑥) − 𝑓(𝑥)_∆𝑥

= lim

∆𝑥→0

3(𝑥+∆𝑥)2−3𝑥2 ∆𝑥

= lim

∆𝑥→0

3(𝑥+∆𝑥)(𝑥+∆𝑥)−3𝑥2

∆𝑥

= lim

∆𝑥→0

3(𝑥2+2𝑥∆𝑥+∆𝑥2)−3𝑥2 ∆𝑥

= lim

∆𝑥→0

3𝑥2_{+6𝑥∆𝑥+3∆𝑥}2_−3𝑥2

∆𝑥

= lim

∆𝑥→0

3𝑥2_−3𝑥2_{+6𝑥∆𝑥+3∆𝑥}2

∆𝑥

= lim

∆𝑥→0

6𝑥∆𝑥+3∆𝑥2 ∆𝑥

= lim

∆𝑥→0

∆𝑥(6𝑥+3∆𝑥) ∆𝑥

= lim

∆𝑥→0 6𝑥+3∆𝑥

1

= lim

∆𝑥→06𝑥 + 3∆𝑥

= 6𝑥 + 3(0) = 6𝑥

Jadi, 𝑓′(𝑥) = 6𝑥

B. Turunan Fungsi Aljabar

Teorema (Penjabaran dari definisi) Jika 𝑓(𝑥) = 𝑎𝑥𝑛, maka 𝑓′(𝑥) = 𝑛𝑎𝑥𝑛−1 Bukti:

Misal: 𝑓(𝑥) = 𝑎𝑥𝑛 𝑓′(𝑥) = 𝑎 lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)𝑛−𝑓(𝑥)𝑛

∆𝑥 ]

= 𝑎 lim

∆𝑥→0[

𝑥𝑛_+𝑛𝑥𝑛−1_∆𝑥+𝑛(𝑛−1)

2 𝑥𝑛−2∆𝑥2+⋯+𝑛𝑥∆𝑥𝑛−1+∆𝑥𝑛−𝑥𝑛

∆𝑥 ]

= 𝑎 lim

∆𝑥→0[

𝑥𝑛_−𝑥𝑛_+𝑛𝑥𝑛−1_∆𝑥+𝑛(𝑛−1)

2 𝑥𝑛−2∆𝑥2+⋯+𝑛𝑥∆𝑥𝑛−1+∆𝑥𝑛

∆𝑥 ]

= 𝑎 lim

∆𝑥→0[

𝑛𝑥𝑛−1_∆𝑥+𝑛(𝑛−1)

2 𝑥𝑛−2∆𝑥2+⋯+𝑛𝑥∆𝑥𝑛−1+∆𝑥𝑛

∆𝑥 ]

= 𝑎 lim

∆𝑥→0[

∆𝑥(𝑛𝑥𝑛−1+𝑛(𝑛−1)₂ 𝑥𝑛−2∆𝑥+⋯+𝑛𝑥∆𝑥𝑛−2+∆𝑥𝑛−1)

∆𝑥 ]

= 𝑎 lim

∆𝑥→0[

𝑛𝑥𝑛−1₊𝑛(𝑛−1)

2 𝑥𝑛−2∆𝑥+⋯+𝑛𝑥∆𝑥𝑛−2+∆𝑥𝑛−1

1 ]

= 𝑎 lim

∆𝑥→0[𝑛𝑥

𝑛−1₊𝑛(𝑛−1)

2 𝑥𝑛−2∆𝑥 + ⋯ + 𝑛𝑥∆𝑥𝑛−2+ ∆𝑥𝑛−1]

= 𝑎 [𝑛𝑥𝑛−1+𝑛(𝑛−1)

2 𝑥𝑛−2(0) + ⋯ + 𝑛𝑥(0)𝑛−2+ (0)𝑛−1]

= 𝑎[𝑛𝑥𝑛−1] = 𝑎𝑛𝑥𝑛−1 = 𝑛𝑎𝑥𝑛−1

3 Ditanya: 𝑓′(𝑥) ?

Jawab: 𝑓(𝑥) = 3𝑥2

𝑓′(𝑥) = 2 ⋅ 3𝑥2−1

= 6𝑥 Jadi, 𝑓′(𝑥) = 6𝑥

C. Operasi Turunan

1) Penjumlahan Turunan

Teorema (Penjabaran dari definisi) Jika 𝑓(𝑥) + 𝑔(𝑥), maka 𝑓′(𝑥) + 𝑔′(𝑥) Bukti:

Misal: 𝑦 = 𝑓(𝑥) + 𝑔(𝑥) 𝑦′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}+{𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 +

𝑔(∆𝑥+𝑥)−𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ] + lim∆𝑥→0[

𝑔(∆𝑥+𝑥)−𝑔(𝑥)

∆𝑥 ]

𝑦′ = 𝑓′(𝑥) + 𝑔′(𝑥)

Contoh: Jika diketahui 𝑓(𝑥) = 3𝑥2dan 𝑔(𝑥) = 2𝑥2, maka tentukan 𝑓′(𝑥) + 𝑔′(𝑥)! Diketahui: 𝑓(𝑥) = 3𝑥2

𝑔(𝑥) = 2𝑥2

Ditanya: 𝑓′(𝑥) + 𝑔′(𝑥) ? Jawab: 𝑓(𝑥) + 𝑔(𝑥) = 3𝑥2+ 2𝑥2

𝑓′(𝑥) + 𝑔′(𝑥) = 2 ⋅ 3𝑥2−1_{+ 2 ⋅ 2𝑥}2−1

= 6𝑥 + 4𝑥 = 10𝑥 Jadi, 𝑓′(𝑥) + 𝑔′(𝑥) = 10𝑥

2) Pengurangan Turunan

Teorema (Penjabaran dari definisi) Jika 𝑓(𝑥) − 𝑔(𝑥), maka 𝑓′(𝑥) − 𝑔′(𝑥) Bukti:

Misal: 𝑦 = 𝑓(𝑥) − 𝑔(𝑥) 𝑦′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}−{𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 −

𝑔(∆𝑥+𝑥)−𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ] − lim∆𝑥→0[

𝑔(∆𝑥+𝑥)−𝑔(𝑥)

∆𝑥 ]

𝑦′ = 𝑓′(𝑥) − 𝑔′(𝑥)

Contoh: Jika diketahui 𝑓(𝑥) = 3𝑥2dan 𝑔(𝑥) = 2𝑥2, maka tentukan 𝑓′(𝑥) − 𝑔′(𝑥)! Diketahui: 𝑓(𝑥) = 3𝑥2

𝑔(𝑥) = 2𝑥2

4 Jawab: 𝑓(𝑥) − 𝑔(𝑥) = 3𝑥2− 2𝑥2

𝑓′(𝑥) − 𝑔′(𝑥) = 2 ⋅ 3𝑥2−1_{− 2 ⋅ 2𝑥}2−1

= 6𝑥 − 4𝑥 = 2𝑥 Jadi, 𝑓′(𝑥) − 𝑔′(𝑥) = 2𝑥

3) Perkalian Turunan

Teorema (Penjabaran dari definisi)

Jika 𝑓(𝑥) ⋅ 𝑔(𝑥), maka [𝑓(𝑥) ⋅ 𝑔(𝑥)]′ = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥) Bukti:

Misal: 𝑦 = 𝑓(𝑥) ⋅ 𝑔(𝑥) 𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(∆𝑥+𝑥)+0−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)+𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)}+{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)}

∆𝑥 +

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)

∆𝑥 +

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 +

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 +

𝑔(𝑥){𝑓(∆𝑥+𝑥)−𝑓(𝑥)}

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ] + lim∆𝑥→0[

𝑔(𝑥){𝑓(∆𝑥+𝑥)−𝑓(𝑥)}

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[𝑓(∆𝑥 + 𝑥)] lim∆𝑥→0[

𝑔(∆𝑥+𝑥)−𝑔(𝑥)

∆𝑥 ] + lim∆𝑥→0[𝑔(𝑥)] lim∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ]

𝑦′ = 𝑓(𝑥)𝑔′(𝑥) + 𝑔(𝑥)𝑓′(𝑥) 𝑦′ = 𝑔(𝑥)𝑓′(𝑥) + 𝑓(𝑥)𝑔′(𝑥) 𝑦′ = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

Contoh: Jika diketahui 𝑓(𝑥) = 3𝑥2dan 𝑔(𝑥) = 2𝑥2, maka tentukan [𝑓(𝑥) ⋅ 𝑔(𝑥)]′! Diketahui: 𝑓(𝑥) = 3𝑥2

𝑔(𝑥) = 2𝑥2

Ditanya: [𝑓(𝑥) ⋅ 𝑔(𝑥)]′ ?

Jawab: [𝑓(𝑥) ⋅ 𝑔(𝑥)]′ = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

= (2 ⋅ 3𝑥2−1)(2𝑥2) + (3𝑥2)(2 ⋅ 2𝑥2−1) = (6𝑥)(2𝑥2) + (3𝑥2)(4𝑥)

= 12𝑥3 + 12𝑥3 = 24𝑥3

5

4) Pembagian Turunan

Teorema (Penjabaran dari definisi)

Jika 𝑓(𝑥)

𝑔(𝑥), maka [ 𝑓(𝑥) 𝑔(𝑥)] ′ =

𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥) [𝑔(𝑥)]2

Bukti:

Misal: 𝑦 =𝑓(𝑥)

𝑔(𝑥)

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)−𝑓(𝑥)𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)

∆𝑥 ]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

𝑔(∆𝑥+𝑥)⋅𝑔(𝑥) ⋅

1 ∆𝑥]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ⋅

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[ 1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)+0−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[ 1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)+𝑓(𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}+{𝑓(𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}−{−𝑓(𝑥)⋅𝑔(𝑥)+𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}−{𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 −

{𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 −

𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 −

𝑓(𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= [ lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 ] − lim∆𝑥→0[

𝑓(𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ]] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦′= [ lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ] lim∆𝑥→0[𝑔(𝑥)] − lim∆𝑥→0[𝑓(𝑥)] lim∆𝑥→0[

𝑔(∆𝑥+𝑥)−𝑔(𝑥) ∆𝑥 ]] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)] 𝑦′ = [𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)] 1

𝑔(𝑥)⋅𝑔(𝑥)

𝑦′ = [𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)] 1

[𝑔(𝑥)]2 𝑦′ =𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)

[𝑔(𝑥)]2

Contoh: Jika diketahui 𝑓(𝑥) = 3𝑥2dan 𝑔(𝑥) = 2𝑥2, maka tentukan [𝑓(𝑥)

𝑔(𝑥)] ′!

Diketahui: 𝑓(𝑥) = 3𝑥2 𝑔(𝑥) = 2𝑥2

Ditanya: [𝑓(𝑥)

𝑔(𝑥)] ′ ?1

Jawab: [𝑓(𝑥)

𝑔(𝑥)] ′ =

𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥) [𝑔(𝑥)]2

=(2⋅3𝑥2−1)(2𝑥2)−(3𝑥2)(2⋅2𝑥2−1)

[2𝑥2_]2 =(6𝑥)(2𝑥2)−(3𝑥2)(4𝑥)

4𝑥4 =12𝑥3−12𝑥3

6 = 0

4𝑥4 = 0 Jadi, [𝑓(𝑥)

𝑔(𝑥)] ′ = 0

D. Teorema L’Hopital

Teorema (Penjabaran dari definisi) Jika 𝑔 ≠ 0 dan 𝑥 ≠ 𝑐, maka

lim

𝑥→𝑐

𝑓(𝑥) 𝑔(𝑥) = lim𝑥→𝑐

𝑓′(𝑥) 𝑔′(𝑥) Bukti:

Misal: 𝑦₁ = 𝑓(𝑥)

𝑔(𝑥)

𝑦₁′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)−𝑔(𝑥)𝑓(𝑥)

∆𝑥 ]

𝑦₁′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)

∆𝑥 ]

𝑦₁′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

𝑔(∆𝑥+𝑥)⋅𝑔(𝑥) ⋅

1 ∆𝑥]

𝑦₁′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ⋅

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[ 1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′= lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)+0−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[ 1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)+𝑓(𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}+{𝑓(𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}−{−𝑓(𝑥)⋅𝑔(𝑥)+𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}−{𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 −

{𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 −

𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 −

𝑓(𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′ = [ lim

∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 ] − lim∆𝑥→0[

𝑓(𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ]] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]

𝑦₁′= [ lim

∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ] lim∆𝑥→0[𝑔(𝑥)] − lim∆𝑥→0[𝑓(𝑥)] lim∆𝑥→0[

𝑔(∆𝑥+𝑥)−𝑔(𝑥) ∆𝑥 ]] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)] 𝑦₁′= [𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)] 1

𝑔(𝑥)⋅𝑔(𝑥)

𝑦₁′= [𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)] 1

[𝑔(𝑥)]2 𝑦₁′= 𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)

[𝑔(𝑥)]2 ⋯ (1)

Misal: 𝑦₂ = 𝑓′(𝑥)

𝑔′(𝑥)

𝑦₂ = lim

∆𝑥→0[

𝑓(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)−𝑓(𝑥)𝑔(𝑥)

∆𝑥 ]

𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)−𝑓(𝑥)𝑔(𝑥)

7 𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥) 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)

∆𝑥 ]]

𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

𝑔(∆𝑥+𝑥)⋅𝑔(𝑥) ⋅

1 ∆𝑥]]

𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ⋅

1

𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[ 1

𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)+0−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[ 1

𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

𝑦₂′ = lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)+𝑓(𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

_𝑦2′_{= lim} ∆𝑥→0[ lim∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}+{𝑓(𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]] 𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}−{−𝑓(𝑥)⋅𝑔(𝑥)+𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]] 𝑦₂′_{= lim}

∆𝑥→0[ lim∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}−{𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]] 𝑦₂′= lim

∆𝑥→0[ lim∆𝑥→0[

{𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)}

∆𝑥 −

{𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)} ∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]] 𝑦₂′ = lim

∆𝑥→0[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)⋅𝑔(𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 −

𝑓(𝑥)⋅𝑔(∆𝑥+𝑥)−𝑓(𝑥)⋅𝑔(𝑥)

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

𝑦₂′ = lim

∆𝑥→0[ lim∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 −

𝑓(𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)}

∆𝑥 ] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

𝑦₂′= lim

∆𝑥→0[[ lim∆𝑥→0[

{𝑓(∆𝑥+𝑥)−𝑓(𝑥)}𝑔(𝑥)

∆𝑥 ] − lim∆𝑥→0[

𝑓(𝑥){𝑔(∆𝑥+𝑥)−𝑔(𝑥)} ∆𝑥 ]] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]]

𝑦₂′= lim ∆𝑥→0[[ lim∆𝑥→0[

𝑓(∆𝑥+𝑥)−𝑓(𝑥)

∆𝑥 ] lim∆𝑥→0[𝑔(𝑥)] − lim∆𝑥→0[𝑓(𝑥)] lim∆𝑥→0[

𝑔(∆𝑥+𝑥)−𝑔(𝑥) ∆𝑥 ]] lim∆𝑥→0[

1 𝑔(∆𝑥+𝑥)⋅𝑔(𝑥)]] 𝑦₂′= lim

∆𝑥→0[[𝑓′(𝑥)𝑔(𝑥)− 𝑓(𝑥)𝑔′(𝑥)]

1 𝑔(𝑥)⋅𝑔(𝑥)]

𝑦₂′= [𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)] 1

𝑔(𝑥)⋅𝑔(𝑥)

𝑦₂′= [𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)] 1

[𝑔(𝑥)]2 𝑦₂′=𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)

[𝑔(𝑥)]2 ⋯ (2)

Dari (1) dan (2)

𝑦1′= 𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)_[𝑔(𝑥)]2 dan 𝑦2′ =

𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)

[𝑔(𝑥)]2 , maka Jika 𝑦₁′= 𝑦₂′, maka lim

𝑥→𝑐 𝑓(𝑥) 𝑔(𝑥)= lim𝑥→𝑐

𝑓′(𝑥) 𝑔′(𝑥)

Contoh: Jika diketahui 𝑓(𝑥) = 𝑥3− 7𝑥2− 8𝑥 − 12 dan 𝑔(𝑥) = 𝑥 − 2 pada saat 𝑥 = 2, maka tentukan [𝑓(𝑥)

𝑔(𝑥)] ′!

Diketahui: 𝑓(𝑥) = 𝑥3 − 7𝑥2 − 8𝑥 − 12 𝑔(𝑥) = 𝑥 − 2

Ditanya: [𝑓(𝑥)

𝑔(𝑥)] ′ ?

Jawab:

𝑓(𝑥) = 𝑥3 _{− 7𝑥}2_{− 8𝑥 − 12}

𝑓′(𝑥) = 3𝑥2_{− 14𝑥− 8}

8 Cara1

[𝑓(𝑥)_𝑔(𝑥)] ′ =𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)_[𝑔(𝑥)]2

= (3𝑥2−14𝑥−8)(𝑥−2)−(𝑥3−7𝑥2−8𝑥−12)(1)

[𝑥−2]2

= (3𝑥3−6𝑥2−14𝑥2+28−8𝑥+16)−(𝑥3−7𝑥2−8𝑥−12)

(𝑥−2)(𝑥−2)

= 3𝑥3−6𝑥2−14𝑥2+28−8𝑥+16−𝑥3+7𝑥2+8𝑥+12

𝑥2_−4𝑥+4

= 3𝑥3−𝑥3−6𝑥2−14𝑥2+7𝑥2−8𝑥+8𝑥+28+16+12

𝑥2_−4𝑥+4

= 2𝑥3−13𝑥2+56

𝑥2_−4𝑥+4 = 2(2)3−13(2)2+56

(2)2_−4(2)+4 = 16−52+56

4−8+4

= 20

0 , tidak terdefinisi

Cara2 lim

𝑥→2 𝑓(𝑥) 𝑔(𝑥)= lim𝑥→2

𝑓′(𝑥) 𝑔′(𝑥)

= lim

𝑥→2

3𝑥2_{−14𝑥−8}

1

= lim

𝑥→23𝑥

2_{− 14𝑥− 8}

= 3(2)2− 14(2) − 8 = 12 − 28 − 8 = −24

Jadi, [𝑓(𝑥)

𝑔(𝑥)] ′

= −24

E. Aplikasi Turunan 1) Gradien

Contoh: Jika diketahui fungsi 𝑓(𝑥) = 𝑥3+ 3𝑥2 dan memotong sumbu 𝑋 di titik (2,0), maka tentukanlah gradiennya!

Diketahui: 𝑓(𝑥) = 𝑥3− 3𝑥2

memotong sumbu 𝑋 di titik (2,0) Ditanya: 𝑚 ?

Jawab: 𝑓(𝑥) = 𝑥3 + 3𝑥2 𝑓′(𝑥) = 3𝑥2_{+ 6𝑥}

𝑓′(2) = 3(2)2_{+ 6(2) = 12 + 12 = 24}

𝑚 = 24

Jadi, 𝑚 = 24

Contoh: Jika diketahui fungsi 𝑓(𝑥) =1

3𝑥3 − 3𝑥2 tentukan koordinat titik singgung

pada gradien −9! Diketahui: 𝑓(𝑥) =1

3𝑥3− 3𝑥2

𝑚 = −9

Ditanya: koordinat titik singgung ? Jawab: 𝑓(𝑥) =1

9

𝑓′(𝑥) = 𝑥2_{− 6𝑥 jika dan hanya jika 𝑚 = 𝑥}2_{− 6𝑥}

−9 = 𝑥2_{− 6𝑥 jika dan hanya jika 𝑥}2 _{− 6𝑥 = −9}

𝑥2_{− 6𝑥 + 9 = 0}

(𝑥 − 3)(𝑥 − 3) = 0 𝑥 − 3 = 0

𝑥 = 3

𝑓(𝑥) =1_{3 𝑥}3_{− 3𝑥}2

𝑦 =1_{3 𝑥}3_{− 3𝑥}2

𝑦 =1₃(3)3_{− 3(3)}2

𝑦 = 9 − 27 = −18

Jadi, koordinat titik singgung (3, −18)

2) Fungsi Naik dan Turun

Untuk setiap 𝑥₁, 𝑥₂ ∈ 𝑆

 Fungsi 𝑓 dikatakan naik jika 𝑥₁ < 𝑥₂, maka 𝑓(𝑥₁) < 𝑓(𝑥₂)  Fungsi 𝑓 dikatakan turun jika 𝑥₁ > 𝑥₂, maka 𝑓(𝑥₁) > 𝑓(𝑥₂)

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3) Titik Stasioner

Syarat mencapai nilai stasioner jika 𝑓′(𝑥) = 0 Contoh: Jika 𝑓(𝑥) =1

3𝑥3− 5

2𝑥2+ 6𝑥, maka tentukanlah titik stasionernya!

Diketahui: 1

3𝑥3− 5

2𝑥2+ 6𝑥

Ditanya: titik stasioner ? Jawab: 𝑓(𝑥) =1

3𝑥3− 5

2𝑥2+ 6𝑥

𝑓′(𝑥) = 𝑥2 − 5𝑥 + 6 jika dan hanya jika 𝑥2_{− 5𝑥 + 6 = 0}

(𝑥 − 2)(𝑥 − 3) = 0 𝑥 − 2 = 0 atau 𝑥 − 3 = 0 𝑥 = 2 atau 𝑥 = 3

Jadi, titik stasionernya adalah (2,0) dan (3,0)

4) Kecepatan dan Percepatan

𝑣 =𝑑𝑥_𝑑𝑡 dan 𝑎 =𝑑𝑣

𝑑𝑡

Keterangan: 𝑥 = jarak 𝑡 = waktu 𝑣 = kecepatan 𝑎 = percepatan

Contoh: Jika diketahui jarak yang ditempuh oleh benda adalah (2𝑡3) meter dan pada saat 𝑡 = 5 detik, maka tentukan kecepatan dan percepatannya! Diketahui: 𝑥 = 2𝑡3

pada saat 𝑡 = 5 detik Ditanya: 𝑣 dan 𝑎 ?

Jawab: 𝑣 =𝑑𝑥_𝑑𝑡 =𝑑(2𝑡3)

𝑑𝑡

= 6𝑡2 = 6(5)2

= 150 meter/detik 𝑎 =𝑑𝑣_𝑑𝑡

=𝑑(6𝑡2)

𝑑𝑡

= 12𝑡 = 12(5)

= 60 meter/detik2

Jadi, kecepatan adalah 150 meter/detik dan percepatan 60 meter/detik2

5) Maksimum dan Minimum

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a) Menentukan nilai fungsi pada batas interval.

b) Menentukan nilai stasioner apabila stationer dicapai pada xdi dalam interval. c) Menentukan nilai minimum dan maksimum berdasarkan hasil dari (a) dan (b). Contoh: Jika 𝑓(𝑥) = −𝑥3+ 6𝑥2 dan pada interval −1 < 𝑥 < 3, maka tentukanlah

nilai maksimum dan minimum! Diketahui: 𝑓(𝑥) = −𝑥3+ 6𝑥2

interval −1 < 𝑥 < 3

Ditanya: nilai maksimum dan minimum ? Jawab:

Nilai fungsi pada batas interval

𝑓(−1) = −(−1)3_{+ 6(−1)}2 _{= 1 + 6 = 7}

𝑓(3) = −(3)3_{+ 6(3)}2 _{= −27 + 54 = 27}

Nilai stasioner 𝑓(𝑥) = −𝑥3_{+ 6𝑥}2

𝑓′(𝑥) = −3𝑥2_{+ 12𝑥 jika dan hanya jika −3𝑥}2 _{+ 12𝑥 = 0}

−3𝑥(𝑥 − 4) = 0

−3𝑥 = 0 atau 𝑥 − 4 = 0 𝑥 = 0 atau 𝑥 = 4

Nilai maksimum dan minimum

𝑓(0) = −(0)3_{+ 6(0)}2 _{= 0 + 0 = 0}

𝑓(4) = −(4)3_{+ 6(4)}2 _{= −64 + 96 = 32}

Sehingga pada interval −1 < 𝑥 < 3 diperoleh 𝑓(−1) = 7

𝑓(3) = 27 𝑓(0) = 0

Jadi, nilai maksimumnya 27 dan nilai minimumnya 0

Contoh: Jika sebuah peluru ditembakkan dari permukaan tanah dengan jarak awal (−𝑡3 _{+ 6𝑡}2_{) meter dengan sudut elevasi 45°, 𝑔 = 10 meter/detik}2_{, dan}

pada saat 𝑡 = 3 detik, maka berapa jarak maksimum yang dicapai peluru! Diketahui: 𝑥₀ = (−𝑡3+ 6𝑡2) meter

𝛼 = 45°

𝑔 = 10 meter/detik2

pada saat 𝑡 = 3 detik Ditanya: 𝑥_𝑚𝑎𝑥 ?

Jawab: 𝑥₀ = −𝑡3+ 6𝑡2

𝑣0 = −3𝑡2+ 12𝑡 = −3(3)2+ 12(3) = −27 + 36 = 9

𝑥𝑚𝑎𝑥 = 𝑣0

2_{sin 2𝛼}

𝑔

=(9)2sin 2(45°)

10

=81 sin 90°

10

=81(1)

10

= 8,1 meter