Definisi
Turunan fungsi \( f \) adalah fungsi \( f' \) yang nilainya di \( c \) adalah
\[f' (c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\]
Jika \( f \) mempunyai turunan di setiap x anggota domain maka :
\[f' (x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
Jika \( y = f(x) \) turunan \( y \) atau turunan \( f \) dinotasikan dengan \( y' \) , atau \( \left( \frac{dy}{dx} \right) dx \ dy \) , atau \( f'(x), \) atau \( \frac{df(x)}{dx} \)
Sifat-Sifat Turunan
Jika \( k \) suatu konstanta, \( f \) dan \( g \) fungsi-fungsi yang terdiferensialkan, \( u \) dan \( v \) fungsi-fungsi dalam \( x \) sehingga \( u = f(x) \) dan \( v = g(x) \) maka berlaku:
Jika \( y = k \ u \) maka \( y' = k(u') \)
Jika \( y = u + v \) maka \( y' = u' + v' \)
Jika \( y = u - v \) maka \( u' - v' \)
Jika \( y = u \ v \) maka \( y' = u' \ v + u \ v' \)
Jika \( y = \frac{u}{v} \) maka \( y' = \frac{ u' \ v - u \ v' }{ v^2 } \)
Latihan Soal
Mencari \( \frac{dy}{dx} \) untuk soal berikut ini
$$ \begin{align} \text{1. } \ y &= ( 3x^4 + 2x^2 + x ) ( x^2 + 7 ) \hspace{100cm} \\ \\ &= 3x^6 + 21x^4 + 2x^4 + 14x^2 + x^3 + 7x \\ \\ &= 3x^6 + 23x^4 + x^3 + 14x^2 + 7x \\ \\ &= 18x^5 + 92x^3 + 3x^2 + 28x + 7 \\ \\ \end{align} $$
$$ \begin{align} \text{2. } \ y &= (x^3 + 3x^2) (4x^2 + 2) \hspace{100cm} \\ \\ &= 4x^5 + 2x^3 + 12x^4 + 6x^2 \\ \\ &= 4x^5 + 12x^4 + 2x^3 + 6x^2 \\ \\ &= 20x^4 + 48x^3 + 6x^2 + 12x \end{align} $$
$$ \begin{align} \text{3. } \ y = \frac{1}{3x^2 + 1}& \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 1; \ u' = 0 \\ \\ v &= 3x^2 + 1; \ v' = 6x \\ \\ \text{Maka, } \\ \\ \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ \\ &= \frac{ 0(3x^2 + 1)-1(6x) }{ (3x^2 + 1)^2 } \\ \\ &= \frac{ -6x }{ (3x^2 + 1)^2 } \end{align} $$
$$ \begin{align} \text{4. } \ y = \frac{2}{5x^2 - 1}& \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 2; \ u' = 0 \\ \\ v &= 5x^2 - 1; \ v' = 10x \\ \\ \text{Maka, } \\ \\ \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ \\ &= \frac{ 0(5x^2 - 1)-2(10x) }{ (5x^2 - 1)^2 } \\ \\ &= \frac{ -20x }{ (5x^2 + 1)^2 } \end{align} $$
$$ \begin{align} \text{5. } \ y = \frac{1}{4x^2 - 3x + 9}& \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 1; \ u' = 0 \\ \\ v &= 4x^2 - 3x + 9; \ v' = 8x - 3 \\ \\ \text{Maka, } \\ \\ \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ \\ &= \frac{ -1 (8x-3) }{ (4x^2 - 3x + 9)^2 } \\ \\ &= \frac{ -8x+3 }{ (4x^2 - 3x + 9)^2 } \end{align} $$
$$ \begin{align} \text{6. } \ y = \frac{x-1}{x+1}& \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= x-1; \ u' = 1 \\ \\ v &= x+1; \ v' = 1 \\ \\ \text{Maka, } \\ \\ \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ \\ &= \frac{ 1(x+1) - (x-1)(1) }{ (x+1)^2 } \\ \\ &= \frac{ x+1 - x+1 }{ (x+1)^2 } \\ \\ &= \frac{2}{(x+1)^2} \end{align} $$
$$ \begin{align} \text{7. } \ y = \frac{2x^2 - 3x + 1}{2x + 1}& \hspace{100cm} \\ \\ \text{Misalkan : } \ u &= 2x^2 - 3x + 1; \ u' = 4x - 3 \\ \\ v &= 2x + 1; \ v' = 2 \\ \\ \text{Maka, } \\ \\ \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ \\ &= \frac{ ( 4x-3 )( 2x+1 ) - ( 2x^2 - 3x + 1 )(2) }{ (2x+1)^2 } \\ \\ &= \frac{ ( 8x^2 + 4x - 6x - 3 - 4x^2 + 6x -2 ) }{ (2x+1)^2 } \\ \\ &= \frac{ 4x^2 + 4x - 5 }{ (2x+1)^2 } \end{align} $$
Menentukan \( f'' (x) \) untuk soal-soal berikut :
$$ \begin{align} \text{1. } \ f(x) &= \sqrt[5]{x^2} + \frac{1}{ \sqrt{x} } \hspace{100cm} \\ \\ &= x^{\frac{2}{5}} + x^{-\frac{1}{2}} \\ \\ f'(x) &= \frac{2}{5}x^{ - \frac{3}{2} } + - \frac{1}{2}x^{ - \frac{3}{2} } \\ \\ &= \frac{2}{5}x^{ - \frac{3}{2} } - \frac{1}{2}x^{ - \frac{3}{2} } \\ \\ f''(x) &= - \frac{6}{25}x^{ - \frac{8}{5} } + \frac{3}{4}x^{ -\frac{5}{2} } \end{align} $$
$$ \begin{align} \text{2. } \ f(x) &= \sqrt{x} \left( 3x + \frac{1}{3x} \right) \left( 3x - \frac{1}{3x} \right) \hspace{100cm} \\ \\ &= \left( 3x^{\frac{3}{2}} + \frac{1}{3}x^{-\frac{1}{2}} \right) \left( 3x - \frac{1}{3}x^{-1} \right) \\ \\ &= 9x^{ \frac{5}{2} } - x^{\frac{1}{2}} + x^{\frac{1}{2}} - \frac{1}{9}x^{ -\frac{3}{2} } \\ \\ &= 9x^{ \frac{5}{2} } - \frac{1}{9}x^{ -\frac{3}{2} } \\ \\ f'(x) &= \frac{45}{2}x^{ \frac{3}{2} } + \frac{3}{18}x^{-\frac{5}{2}} \\ \\ f''(x) &= \frac{135}{4}x^{\frac{1}{2}} - \frac{15}{36}x^{ -\frac{7}{2} } \end{align} $$
$$ \begin{align} \text{3. } \ f(x) &= (5x^2 - 1) (x^2 + 4x -2) \hspace{100cm} \\ \\ &= 5x^4 + 20x^3 - 10x^2 - x^2 - 4x + 2 \\ \\ &= 5x^4 + 20x^3 - 11x^2 - 4x + 2 \\ \\ f'(x) &= 20x^3 + 60x^2 - 22x - 4 \\ \\ f''(x) &= 60x^2 + 120x - 22 \end{align} $$