Konsep
Untuk menentukan turunan \( y = (3x^4 + 7 − 8)^9 \) dengan cara mengalikan bersama kesembilan faktor \( y = (3x^4 + 7 − 8) \) kemudian mencari turunan polinom berderajat 36 tentulah sangat melelahkan. Cara yang mudah untuk menentukan turunan \( y = (3x^4 + 7 −8)^9 \) adalah dengan menggunakan aturan rantai.
Misalkan \( y = f(u) \) dan \( u = g(x) \) menentukan fungsi komposisi yang dirumuskan dengan \( y = f(g(x)) = (f \circ g)(x) \). Jika \( g \) terdiferensialkan di \( x \) dan \( f \) terdiferensialkan di \( u = g(x) \) maka \( y = (f \circ g)(x) \) terdiferensialkan di \( x \) dan
$$ \begin{align} y' &= (f \circ g)' (x) \\ \\ &= f' (g(x)) g'(x) \end{align} $$
atau
$$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} \hspace{100cm} $$
Fungsí komposisi dapat diperluas menjadi komposisi 3 fungsi, 4 fungsi dan seterusnya.
$$ \begin{align} \text{Jika } \ y &= f(u) \\ \\ u &= g(v) \\ \\ v &= h(x)\\ \\ \text{yakni } \ y &= (f \circ g \circ h)(x) \\ \\ \text{maka} \ \frac{dy}{dx} &= \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx} \end{align} $$
Dengan aturan rantai menentukan \( \frac{dy}{dx} \) untuk persoalan berikut ini
$$ \begin{align} \text{1. } \ y = (2 - 9x&)^{15} \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 2-9x \ ; \ u' = -9 \\ \\ y &= u^{15} \ ; \ y' = 15u^{14} \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} = 15u^{14} (-9) \\ \\ &= -135(2-9x)^{14} \end{align} $$
$$ \begin{align} \text{2. } \ y = (5x^2 + 2x& - 8)^5 \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 5x^2 + 2x - 8 \ ; \ u' = 10x + 2 \\ \\ y &= u^{5} \ ; \ y' = 5u^{4} \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} \\ \\ &= (10x+2)(5u^4) \\ \\ &= (10x+2) ( 5 (5x^2+2x-8)^4 ) \\ \\ &= (20x + 10) (5x^2 + 2x - 8)^4 \end{align} $$
$$ \begin{align} \text{3. } \ y = \frac{1}{ (4x^2 - 3x + 9) }& = (4x^2-3x+9)^{-9} \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 4x^2 - 3x + 9, \ \quad \frac{du}{dx} = 8x-3 \\ \\ y &= 4^{-9} , \ \qquad \qquad \quad \frac{dy}{du} = -9u^{-10} \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} \\ \\ &= (8x-3) \left( -9 (4x^2 - 3x +9) \right)^{-10} \\ \\ &= (-72x+27)(4x^2-3x+9)^{-10} \end{align} $$
$$ \begin{align} \text{4. } \ y = sin(3x^2 - & 11x) \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 3x^2 - 11x \ \quad \frac{du}{dx} = 6x+11 \\ \\ y &= \sin \ u \ \qquad \quad \frac{dy}{du} = \cos u \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} = (6x-11) \left( \cos (3x^2 + 11x) \right) \end{align} $$
$$ \begin{align} \text{5. } \ y = \cos (3x^4 + &11x) \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 3x^4 + 11x \ \quad \frac{du}{dx} = 12x^3-11 \\ \\ y &= \cos \ u \ \qquad \quad \frac{dy}{du} = -\sin u \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} = 12x^3-11 \left( - \sin (3x^4 -11x) \right) \end{align} $$
$$ \begin{align} \text{6. } \ y = \cos & \left( \frac{x-1}{x+1} \right)^4 \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= \frac{x-1}{x+1} \ \quad \quad \quad \frac{du}{dx} = \frac{1(x+1) - (x-1)(1)}{(x+1)^2} \\ \\ & \qquad \qquad \qquad \quad \quad \quad= \frac{x+1-x+1}{(x+1)^2} \\ \\ & \qquad \qquad \qquad \quad \quad \quad= \frac{2}{(x+1)^2} \\ \\ y &= u^4 \ \qquad \quad \quad \quad \frac{dy}{du} = 4u^3 \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} = \frac{2}{(x+1)^2} \left( 4 (u^3) \right) \\ \\ &= \frac{8u^3}{(x+1)^2} \\ \\ &= \frac{ 8 \left( \frac{x-1}{x+1} \right) }{ (x+1)^2 } \\ \\ &= \frac{8x - 8}{x+1} \cdot \frac{1}{(x+1)^2} \\ \\ &= \frac{8x-8}{(x+1)^3} \\ \\ &= 8 \left( \frac{x-1}{(x+1)^3} \right) \\ \\ \end{align} $$
$$ \begin{align} \text{7. } \ y = \sin & \left( \frac{3x-1}{2x+5} \right) \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= \frac{3x-1}{2x+5} \ \quad \quad \quad \frac{du}{dx} = \frac{ 3(2x+5) - (3x-1)(2) }{ (2x+5)^2 } \\ \\ & \qquad \qquad \qquad \quad \quad \quad \quad= \frac{ 6x+15-(6x-2) }{(2x+5)^2} \\ \\ & \qquad \qquad \qquad \quad \quad \quad \quad= \frac{17}{(2x+5)^2} \\ \\ y &= \sin u \ \qquad \quad \quad \frac{dy}{du} = \cos u \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} = \frac{17}{(2x+5)^2} \cos u \\ \\ &= \frac{ 17 \cos \left( \frac{3x-1}{2x+5} \right) }{ (2x+5)^2 } \end{align} $$
$$ \begin{align} \text{8. } \ y = \cos & \left( \frac{ x^2 - 1 }{ x + 4 } \right) \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= \frac{ x^2 - 1 }{ x + 4 } \ \quad \quad \quad \frac{du}{dx} = \frac{ 2x(x+4) - 1 (x^2-1) }{ (x+4)^2 } \\ \\ & \qquad \qquad \qquad \quad \quad \quad \ = \frac{ 2x^2 + 8x -x^2 + 1 }{(x+4)^2} \\ \\ & \qquad \qquad \qquad \quad \quad \quad \ = \frac{x^2 + 8x + 1}{(x+4)^2} \\ \\ y &= \cos u \ \qquad \quad \quad \frac{dy}{du} = - \sin u \\ \\ \text{Maka : } \ \\ \\ \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} \\ \\ &= - \sin \left( \frac{ x^2 - 1 }{ x + 4 } \right) \cdot \left( \frac{ x^2 + 8x + 1 }{ (x+4)^3 } \right) \\ \\ &= - \sin \left( \frac{ (x^2-1)(x^2+8x+1) }{ (x+4)^3 } \right) \end{align} $$