Turunan adalah pengukuran terhadap bagaimana nilai suatu fungsi berubah seiring perubahan nilai inputnya. Secara geometris, turunan merepresentasikan kemiringan (gradien) garis singgung pada suatu titik di kurva.
Turunan didefinisikan menggunakan limit. Notasi Δ x \Delta x Δ x x x x
f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}
f ′ ( x ) = Δ x → 0 lim Δ x f ( x + Δ x ) − f ( x ) Notasi Dibaca Keterangan f ′ ( x ) f'(x) f ′ ( x ) f aksen x Notasi Lagrange d y d x \frac{dy}{dx} d x d y dy per dx Notasi Leibniz d f d x \frac{df}{dx} d x df df per dx Notasi Leibniz D f ( x ) Df(x) D f ( x ) D dari f(x) Notasi Euler
Turunan f ′ ( a ) f'(a) f ′ ( a ) y = f ( x ) y = f(x) y = f ( x ) x = a x = a x = a
Jika f ( x ) = k f(x) = k f ( x ) = k
f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0
Contoh: f ( x ) = 5 ⇒ f ′ ( x ) = 0 f(x) = 5 \quad \Rightarrow \quad f'(x) = 0 f ( x ) = 5 ⇒ f ′ ( x ) = 0
Jika f ( x ) = a x n f(x) = ax^n f ( x ) = a x n
f ′ ( x ) = n ⋅ a ⋅ x n − 1 f'(x) = n \cdot a \cdot x^{n-1} f ′ ( x ) = n ⋅ a ⋅ x n − 1
Cara Cepat: "Pangkat turun ke depan dikali koefisien, lalu pangkatnya dikurangi satu."
Contoh:
f ( x ) = x 3 ⇒ f ′ ( x ) = 3 x 2 f(x) = x^3 \quad \Rightarrow \quad f'(x) = 3x^2 f ( x ) = x 3 ⇒ f ′ ( x ) = 3 x 2 f ( x ) = 5 x 4 ⇒ f ′ ( x ) = 20 x 3 f(x) = 5x^4 \quad \Rightarrow \quad f'(x) = 20x^3 f ( x ) = 5 x 4 ⇒ f ′ ( x ) = 20 x 3 f ( x ) = x ⇒ f ′ ( x ) = 1 f(x) = x \quad \Rightarrow \quad f'(x) = 1 f ( x ) = x ⇒ f ′ ( x ) = 1
Turunan dari penjumlahan/pengurangan fungsi adalah jumlah/selisih dari turunannya.
( f ± g ) ′ = f ′ ( x ) ± g ′ ( x ) (f \pm g)' = f'(x) \pm g'(x) ( f ± g ) ′ = f ′ ( x ) ± g ′ ( x )
Soal: Tentukan turunan dari y = 2 x 4 − 6 x + 15 y = 2x^4 - 6x + 15 y = 2 x 4 − 6 x + 15
Penyelesaian:
Turunkan setiap suku secara terpisah:
Suku Proses Hasil 2 x 4 2x^4 2 x 4 4 ⋅ 2 ⋅ x 4 − 1 4 \cdot 2 \cdot x^{4-1} 4 ⋅ 2 ⋅ x 4 − 1 8 x 3 8x^3 8 x 3 − 6 x -6x − 6 x 1 ⋅ ( − 6 ) ⋅ x 1 − 1 1 \cdot (-6) \cdot x^{1-1} 1 ⋅ ( − 6 ) ⋅ x 1 − 1 − 6 -6 − 6 15 15 15 Konstanta 0 0 0
Hasil akhir: y ′ = 8 x 3 − 6 y' = 8x^3 - 6 y ′ = 8 x 3 − 6
Soal: Tentukan turunan dari f ( x ) = x + 1 x 2 f(x) = \sqrt{x} + \frac{1}{x^2} f ( x ) = x + x 2 1
Penyelesaian:
Langkah 1: Ubah ke bentuk pangkat
f ( x ) = x 1 / 2 + x − 2 f(x) = x^{1/2} + x^{-2} f ( x ) = x 1/2 + x − 2
Langkah 2: Turunkan
f ′ ( x ) = 1 2 x 1 / 2 − 1 + ( − 2 ) x − 2 − 1 = 1 2 x − 1 / 2 − 2 x − 3 = 1 2 x − 2 x 3 \begin{aligned}
f'(x) &= \frac{1}{2}x^{1/2-1} + (-2)x^{-2-1} \\[8pt]
&= \frac{1}{2}x^{-1/2} - 2x^{-3} \\[8pt]
&= \frac{1}{2\sqrt{x}} - \frac{2}{x^3}
\end{aligned}
f ′ ( x ) = 2 1 x 1/2 − 1 + ( − 2 ) x − 2 − 1 = 2 1 x − 1/2 − 2 x − 3 = 2 x 1 − x 3 2 Digunakan untuk fungsi komposisi (fungsi di dalam fungsi), seperti y = ( u ( x ) ) n y = (u(x))^n y = ( u ( x ) ) n
d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) ⋅ g ′ ( x ) Atau dalam bentuk yang lebih sederhana:
Jika y = u n maka y ′ = n ⋅ u n − 1 ⋅ u ′ \text{Jika } y = u^n \quad \text{maka} \quad y' = n \cdot u^{n-1} \cdot u' Jika y = u n maka y ′ = n ⋅ u n − 1 ⋅ u ′
Konsep: Turunkan fungsi luarnya (anggap fungsi dalam sebagai variabel biasa), lalu kalikan dengan turunan fungsi dalamnya.
Contoh: Tentukan turunan y = ( 3 x 2 − 5 ) 4 y = (3x^2 - 5)^4 y = ( 3 x 2 − 5 ) 4
Penyelesaian:
Fungsi dalam: u = 3 x 2 − 5 ⇒ u ′ = 6 x u = 3x^2 - 5 \quad \Rightarrow \quad u' = 6x u = 3 x 2 − 5 ⇒ u ′ = 6 x
Fungsi luar: ( u ) 4 (u)^4 ( u ) 4
y ′ = 4 ( 3 x 2 − 5 ) 4 − 1 ⋅ ( 6 x ) = 4 ⋅ 6 x ⋅ ( 3 x 2 − 5 ) 3 = 24 x ( 3 x 2 − 5 ) 3 \begin{aligned}
y' &= 4(3x^2 - 5)^{4-1} \cdot (6x) \\[8pt]
&= 4 \cdot 6x \cdot (3x^2 - 5)^3 \\[8pt]
&= 24x(3x^2 - 5)^3
\end{aligned} y ′ = 4 ( 3 x 2 − 5 ) 4 − 1 ⋅ ( 6 x ) = 4 ⋅ 6 x ⋅ ( 3 x 2 − 5 ) 3 = 24 x ( 3 x 2 − 5 ) 3 Digunakan jika f ( x ) f(x) f ( x )
Jika f ( x ) = u ⋅ v maka f ′ ( x ) = u ′ ⋅ v + u ⋅ v ′ \text{Jika } f(x) = u \cdot v \quad \text{maka} \quad f'(x) = u' \cdot v + u \cdot v' Jika f ( x ) = u ⋅ v maka f ′ ( x ) = u ′ ⋅ v + u ⋅ v ′
Hafalan: "Turunan depan kali belakang, ditambah depan kali turunan belakang."
Contoh: Tentukan turunan y = ( 2 x + 3 ) ( x 2 ) y = (2x + 3)(x^2) y = ( 2 x + 3 ) ( x 2 )
Penyelesaian:
Misal u = 2 x + 3 ⇒ u ′ = 2 u = 2x + 3 \quad \Rightarrow \quad u' = 2 u = 2 x + 3 ⇒ u ′ = 2
Misal v = x 2 ⇒ v ′ = 2 x v = x^2 \quad \Rightarrow \quad v' = 2x v = x 2 ⇒ v ′ = 2 x
y ′ = u ′ ⋅ v + u ⋅ v ′ = ( 2 ) ( x 2 ) + ( 2 x + 3 ) ( 2 x ) = 2 x 2 + 4 x 2 + 6 x = 6 x 2 + 6 x \begin{aligned}
y' &= u' \cdot v + u \cdot v' \\[8pt]
&= (2)(x^2) + (2x+3)(2x) \\[8pt]
&= 2x^2 + 4x^2 + 6x \\[8pt]
&= 6x^2 + 6x
\end{aligned} y ′ = u ′ ⋅ v + u ⋅ v ′ = ( 2 ) ( x 2 ) + ( 2 x + 3 ) ( 2 x ) = 2 x 2 + 4 x 2 + 6 x = 6 x 2 + 6 x Digunakan jika f ( x ) f(x) f ( x )
Jika f ( x ) = a b maka f ′ ( x ) = a ′ ⋅ b − a ⋅ b ′ b 2 \text{Jika } f(x) = \frac{a}{b} \quad \text{maka} \quad f'(x) = \frac{a' \cdot b - a \cdot b'}{b^2}
Jika f ( x ) = b a maka f ′ ( x ) = b 2 a ′ ⋅ b − a ⋅ b ′
**Hati-hati! ** Tandanya adalah kurang (-) , dan urutannya tidak boleh terbalik.
Hafalan: "Turunan atas kali bawah, dikurang atas kali turunan bawah, dibagi bawah kuadrat."
Contoh: Tentukan turunan y = x 2 x + 1 y = \frac{x^2}{x+1} y = x + 1 x 2
Penyelesaian:
Atas: a = x 2 ⇒ a ′ = 2 x a = x^2 \quad \Rightarrow \quad a' = 2x a = x 2 ⇒ a ′ = 2 x
Bawah: b = x + 1 ⇒ b ′ = 1 b = x+1 \quad \Rightarrow \quad b' = 1 b = x + 1 ⇒ b ′ = 1
y ′ = u ′ ⋅ v − u ⋅ v ′ v 2 = ( 2 x ) ( x + 1 ) − ( x 2 ) ( 1 ) ( x + 1 ) 2 = 2 x 2 + 2 x − x 2 ( x + 1 ) 2 = x 2 + 2 x ( x + 1 ) 2 \begin{aligned}
y' &= \frac{u' \cdot v - u \cdot v'}{v^2} \\[8pt]
&= \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2} \\[8pt]
&= \frac{2x^2 + 2x - x^2}{(x+1)^2} \\[8pt]
&= \frac{x^2 + 2x}{(x+1)^2}
\end{aligned}
y ′ = v 2 u ′ ⋅ v − u ⋅ v ′ = ( x + 1 ) 2 ( 2 x ) ( x + 1 ) − ( x 2 ) ( 1 ) = ( x + 1 ) 2 2 x 2 + 2 x − x 2 = ( x + 1 ) 2 x 2 + 2 x Aturan Rumus Konstanta d d x ( k ) = 0 \frac{d}{dx}(k) = 0 d x d ( k ) = 0 Pangkat d d x ( x n ) = n x n − 1 \frac{d}{dx}(x^n) = nx^{n-1} d x d ( x n ) = n x n − 1 Perkalian ( a b ) ′ = a ′ b + a b ′ (ab)' = a'b + ab' ( ab ) ′ = a ′ b + a b ′ Pembagian ( a b ) ′ = a ′ b − a b ′ b 2 \left(\frac{a}{b}\right)' = \frac{a'b - ab'}{b^2} ( b a ) ′ = b 2 a ′ b − a b ′ Rantai d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) ⋅ g ′ ( x )
Ubah bentuk terlebih dahulu! Akar dan pecahan sebaiknya diubah ke bentuk pangkat sebelum diturunkan.
Identifikasi aturan yang tepat! Perhatikan apakah fungsi berupa perkalian, pembagian, atau komposisi.
Hati-hati dengan tanda! Terutama pada aturan pembagian yang menggunakan pengurangan.