$$\bbox[yellow, 5px, border: 2px solid red] {y = a \cdot x^n \rightarrow y’ = a \cdot n \cdot x^{n\:-\:1}}$$
\(y’ = \dfrac{dy}{dx} = \text{ turunan pertama y}\)
Contoh 01
Tentukan turunan pertama dari \(y = 3x^5\)
Penyelesaian:
\(y’ = 3 \cdot 5 \cdot x^{5\:-\:1}\)
\(y’ = 15x^4\)
Contoh 02
Tentukan turunan pertama dari \(y = 25x\)
Penyelesaian:
\(y’ = 25 \cdot 1 \cdot x^{1\:-\:1}\)
\(y’ = 25x^0\)
\(y’ = 25\cdot 1 = 25\)
Contoh 03
Tentukan turunan pertama dari \(y = 25\)
Penyelesaian:
\(y = 25 x^{0}\)
\(y’ = 25 \cdot 0 \cdot x^{0\:-\:1}\)
\(y’ = 0\)
Contoh 04
Tentukan turunan pertama dari \(y = x\sqrt{x}\)
Penyelesaian:
\(y = x\sqrt{x}\)
\(y = x^1\cdot x^{\frac{1}{2}}\)
\(y = x^{\frac{3}{2}}\)
\(y’ = \dfrac{3}{2} \cdot x^{\frac{3}{2}\:-\:1}\)
\(y’ = \dfrac{3}{2} \cdot x^{\frac{1}{2}}\)
\(y’ = \dfrac{3}{2} \sqrt{x}\)
Aturan Berantai
$$\bbox[yellow, 5px, border: 2px solid red] {\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}}$$
Contoh 01
Jika \(y = u^5\) dan \(u = 2x + 1\), tentukan \(\dfrac{dy}{dx}\)
Penyelesaian:
\(y = u^5 \rightarrow \color{blue} \dfrac{dy}{du} = 5\cdot u^4\)
\(u = 2x + 1 \rightarrow \color{blue} \dfrac{du}{dx} = 2\)
\(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)
\(\dfrac{dy}{dx} = 5\cdot u^4 \times 2\)
\(\dfrac{dy}{dx} = 10(2x + 1)^4\)
Contoh 02
Tentukan turunan pertama dari \(y = (5x + 2)^{10}\)
Penyelesaian:
Misal \(u = 5x + 2\), maka \(y = u^{10}\)
\(y = u^{10} \rightarrow \color{blue} \dfrac{dy}{du} = 10\cdot u^{9}\)
\(u = 5x + 2 \rightarrow \color{blue} \dfrac{du}{dx} = 5\)
\(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)
\(\dfrac{dy}{dx} = 10\cdot u^{9} \times 5\)
\(\dfrac{dy}{dx} = 50(5x + 2)^{9}\)
Bentuk Perkalian dan Pembagian
Turunan Pertama Bentuk Perkalian Fungsi
$$\bbox[yellow, 5px, border: 2px solid red] {y = u \cdot v \rightarrow y’ = u’ \cdot v + u \cdot v’}$$
$$\bbox[yellow, 5px, border: 2px solid red] {y = u \cdot v \cdot w \rightarrow y’ = u’ \cdot v \cdot w + u \cdot v’ \cdot w + u \cdot v \cdot w’}$$
Contoh:
Tentukan turunan pertama dari \(y = (2x + 1)^5 \cdot (x \:-\:3)^2\)
Penyelesaian:
Misal \(u = (2x + 1)^5\) dan \(v = (x \:-\:3)^2\)
\(u = (2x + 1)^5\)
\(u’ = 2 \cdot 5 \cdot (2x + 1)^{5\:-\:1}\)
\(u’ = 10 (2x + 1)^{4}\)
\(v = (x \:-\:3)^2\)
\(v’ = 2 \cdot (x\:-\:3)^{2\:-\:1}\)
\(v’ = 2(x\:-\:3)\)
\(y’ = u’ \cdot v + u \cdot v’\)
\(y’ = 10 (2x + 1)^{4}\cdot (x \:-\:3)^2 + (2x + 1)^5 \cdot 2(x\:-\:3)\)
\(y’ = 2\cdot (2x + 1)^4 \cdot (x\:-\:3) \cdot [5(x\:-\:3) + (2x + 1)]\)
\(y’ = 2\cdot (2x + 1)^4 \cdot (x\:-\:3) \cdot (7x\:-\:14)\)
Turunan Pertama Bentuk Pembagian Fungsi
$$\bbox[yellow, 5px, border: 2px solid red] {y = \dfrac{u}{v} \rightarrow y’ = \dfrac{u’ \cdot v \:-\:u \cdot v’}{v^2}}$$
Contoh:
Tentukan turunan pertama dari \(y = \dfrac{2x + 1}{5x\:-\:1}\)
Penyelesaian:
Misal \(u = 2x + 1\) dan \(v = 5x\:-\:1\)
\(u = 2x + 1 \rightarrow \color{blue} u’ = 2\)
\(v = 5x\:-\:1 \rightarrow \color{blue} v’ = 5\)
\(y’ = \dfrac{u’ \cdot v \:-\:u \cdot v’}{v^2}\)
\(y’ = \dfrac{2\cdot (5x\:-\:1) \:-\:(2x + 1) \cdot 5}{(5x\:-\:1)^2}\)
\(y’ = \dfrac{10x\:-\:2\:-\:10x\:-\:5}{(5x\:-\:1)^2}\)
\(y’ = \dfrac{-7}{(5x\:-\:1)^2}\)
LATIHAN SOAL
Soal 01
Turunan pertama dari \(y = 3x^5 \:-\:4x^2 + 1\) adalah…
(A) \(y’ = 15x^4 \:-\:8x\)
(B) \(y’ = 15x^4 \:-\:8x + 1\)
(C) \(y’ = 15x^4 + 8x\)
(D) \(y’ = 8x^4 + 8x^2 + 1\)
(E) \(y’ = 3x^4 \:-\:4x + 1\)
Soal 02
Turunan pertama dari \(y = 2\sqrt{x} + \dfrac{1}{x}\) adalah…
(A) \(y’ = \dfrac{1}{\sqrt{x^3}}\:-\:\dfrac{2}{x}\)
(B) \(y’ = \dfrac{2}{\sqrt{x}}\:-\:\dfrac{1}{x}\)
(C) \(y’ = \dfrac{1}{\sqrt{x}}\:-\:\dfrac{1}{x^2}\)
(D) \(y’ = \dfrac{2}{\sqrt{x}}\:-\:\dfrac{1}{x^2}\)
(E) \(y’ = \dfrac{1}{\sqrt{x}} + \dfrac{1}{x^2}\)
Soal 03
Turunan pertama dari \(y = \dfrac{x^2 + 2x\:-\:1}{\sqrt{x}}\) adalah…
(A) \(y’ = \dfrac{1}{2}\sqrt{x} + \dfrac{1}{\sqrt{x}} \:-\: \dfrac{1}{2x\sqrt{x}}\)
(B) \(y’ = \dfrac{3}{2}\sqrt{x} + \dfrac{1}{\sqrt{x^2}} + \dfrac{1}{x\sqrt{x}}\)
(C) \(y’ = \dfrac{3}{2}\sqrt{x} \:-\: \dfrac{1}{\sqrt{x}} + \dfrac{1}{2x\sqrt{x}}\)
(D) \(y’ = \dfrac{1}{2}\sqrt{x} + \dfrac{1}{\sqrt{x}} + \dfrac{1}{2x\sqrt{x}}\)
(E) \(y’ = \dfrac{3}{2}\sqrt{x} + \dfrac{1}{\sqrt{x}} + \dfrac{1}{2x\sqrt{x}}\)
Soal 04
Turunan pertama dari \(y = \dfrac{1}{(2x\:-\:3)^5}\) adalah…
(A) \(y’ = -\dfrac{5}{(2x + 3)^6}\)
(B) \(y’ = -\dfrac{2}{(2x\:-\:3)^6}\)
(C) \(y’ = -\dfrac{10}{(2x\:-\:3)^5}\)
(D) \(y’ = -\dfrac{8}{(2x\:-\:3)^6}\)
(E) \(y’ = -\dfrac{10}{(2x\:-\:3)^6}\)
Soal 05
Turunan pertama dari \(y = (5x + 2)^2 \cdot (5x\:-\:2)^2\) adalah…
(A) \(y’ = 2000x^3 + 200x\)
(B) \(y’ = 2000x^3\:-\:400x\)
(C) \(y’ = 2500x^3\:-\:400x\)
(D) \(y’ = 2500x^3 + 400x\)
(E) \(y’ = 2500x^3\:-\:800x\)